Probability

No. of rusted screws = $\frac{1}{10} \times 600$ = 60,

No. of good screws = 600 - 60 = 540,

Let E₁ be the event of taking out a good screw, then number of favourable outcomes to E₁ = 540

$P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{540}{600} = \frac{9}{10}.$

Hence, the probability that the screw taken out is a good screw is $\frac{9}{10}$

Total no. of tickets = 995 + 5 = 1000

Let E₁ be the event of getting a prized ticket, then number of favourable outcomes to E₁ = 5

$P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{5}{1000} = \frac{1}{200}.$

Hence, the probability of winning a prize is $\frac{1}{200}$.

No. of defective pens = 12

No. of good pens = 132

Total no. of pens = 12 + 132 = 144.

Let E₁ be the event of getting a good pen, then number of favourable outcomes to E₁ = 132

$P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{132}{144} = \frac{11}{12}.$

Hence, the probability of taking out a good pen is $\frac{11}{12}$.

Let P(E) be the probability of Sania's winning and P(E') be the probability of Sania losing or the probability of Sonali winning.

∴ P(E) + P(E') = 1

⇒ 0.69 + P(E') = 1
⇒ P(E') = 1 - 0.69
⇒ P(E') = 0.31

Hence, the probability of Sonali winning is 0.31

(i) No. of red balls = 3
    No. of black balls = 5.

Total no. of balls = 3 + 5 = 8.

Let E₁ be the event of drawing a red ball, then number of favourable outcomes to E₁ = 3

$P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{3}{8}.$

Hence, the probability of drawing out a red ball is $\frac{3}{8}$.

(ii) Since, there are only two coloured balls red and black. Hence, the probability of not drawing a red ball = probability of drawing a black ball.

Let E₂ be the event of drawing a black ball, then number of favourable outcomes to E₂ = 5

$P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{5}{8}.$

Hence, the probability of drawing out a not red ball is $\frac{5}{8}$.

Vowels in 'TRIANGLE' = 'I', 'A', 'E'.

No. of letters in 'TRIANGLE' = 8.

Let E₁ be the event of choosing a vowel, then number of favourable outcomes to E₁ = 3

$P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{3}{8}.$

Hence, the probability of choosing a vowel is $\frac{3}{8}$.

Total alphabets = 26

No. of vowels = 5

No. of consonants = 21

Let E₁ be the event of choosing a consonant, then number of favourable outcomes to E₁ = 21

$P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{21}{26}.$

Hence, the probability of choosing a consonant is $\frac{21}{26}$.

Total number of balls = 5 + 2 + 3 = 10

So, the total number of possible outcomes = 10

There are 2 red balls.

∴ Number of favourable outcomes = 2

P(getting a red ball) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{2}{10} = \frac{1}{5}$.

Hence, probability of getting a red ball = $\frac{1}{5}$.

(i) No. of black marbles = 5 and Total marbles = 7 + 8 + 5 = 20.

Let E₁ be the event of choosing a black marble, then number of favourable outcomes to E₁ = 5

$P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{5}{20} = \frac{1}{4}.$

Hence, the probability of choosing a black marble is $\frac{1}{4}$.

(ii) Total no. of black and blue marbles = 7 + 5 = 12.

Let E₂ be the event of choosing a black or blue marble, then number of favourable outcomes to E₂ = 12

$P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{12}{20} = \frac{3}{5}.$

Hence, the probability of choosing a black or blue marble is $\frac{3}{5}$.

(iii) Total no. of blue and white marbles = 7 + 8 = 15.

Let E₃ be the event of choosing a white or blue marble, then number of favourable outcomes to E₃ = 15

$P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{15}{20} = \frac{3}{4}.$

Hence, the probability of choosing a not black marble is $\frac{3}{4}$.

(iv) Let E₄ be the event of choosing a green marble, then number of favourable outcomes to E₄ = 0, as there is no green marble in the box.

$P(E_4) = \frac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \frac{0}{20} = 0.$

Hence, the probability of choosing a green marble is 0.

(i) No. of white balls = 8 and total no. of balls = 6 + 8 + 5 + 3 = 22.

Let E₁ be the event of choosing a white ball, then number of favourable outcomes to E₁ = 8

$P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{8}{22} = \frac{4}{11}.$

Hence, the probability of drawing a white ball is $\frac{4}{11}$.

(ii) Total no. of red and black balls = 6 + 3 = 9.

Let E₂ be the event of choosing a red or black ball, then number of favourable outcomes to E₂ = 9

$P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{9}{22}.$

Hence, the probability of drawing a red or black ball is $\frac{9}{22}$.

(iii) Probability of not drawing a green ball means probability of drawing any other colour ball.

Total no. of red, black and white balls = 6 + 3 + 8 = 17.

Let E₃ be the event of choosing a red, black or white ball, then number of favourable outcomes to E₃ = 17

$P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{17}{22}.$

Hence, the probability of drawing a not green ball is $\frac{17}{22}$.

(iv) Probability of not drawing a white or black ball means probability of drawing red or green ball.

Let E₄ be the event of choosing a red or green ball, then number of favourable outcomes to E₄ = 11

$P(E_4) = \frac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \frac{11}{22} = \frac{1}{2}.$

Hence, the probability of drawing neither white nor black ball is $\frac{1}{2}$.

(i) No. of good shirts = 88
    Total no. of shirts = 100.

Let E₁ be the event of choosing a good shirt, then number of favourable outcomes to E₁ = 88

$P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{88}{100} = \frac{22}{25}.$

Hence, the probability that shirt is acceptable to Peter is $\frac{22}{25}$.

(ii) No. of good and minor defective shirts = 88 + 8 = 96
    Total no. of shirts = 100.

Let E₂ be the event of choosing a good or minor defective shirt, then number of favourable outcomes to E₂ = 96

$P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{96}{100} = \frac{24}{25}.$

Hence, the probability that shirt is acceptable to Salim is $\frac{24}{25}$.

(i) Dice is thrown once
    Sample space = {1, 2, 3, 4, 5, 6} which has 6 likely outcomes.

Let E₁ be the event of getting a even number, then number of favourable outcomes to E₁ = 3

$P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{3}{6} = \frac{1}{2}.$

Hence, the probability of getting a even no. is $\frac{1}{2}$.

(ii) Let E₂ be the event of getting a number greater than 2, then number of favourable outcomes to E₂ = 4

$P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{4}{6} = \frac{2}{3}.$

Hence, the probability of getting a no. greater than 2 is $\frac{2}{3}$.

In a single throw of die,

Sample space = {1, 2, 3, 4, 5, 6}.

(i) Let E be the event of getting an odd number, then

E = {1, 3, 5}.

∴ The number of favourable outcomes to the event E = 3.

$\therefore P(E) = \frac{\text{No. of favourable outcomes to } E}{\text{Total no. of possible outcomes}} = \frac{3}{6} = \frac{1}{2}.$

Hence, the probability of getting an odd number is $\frac{1}{2}$.

(ii) Let E₁ be the event of getting a number less than 5, then

E₁ = {1, 2, 3, 4}.

∴ The number of favourable outcomes to the event E₁ = 4.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{4}{6} = \frac{2}{3}.$

Hence, the probability of getting a number less than 5 is $\frac{2}{3}$.

(iii) Let E₂ be the event of getting a number greater than 5, then

E₂ = {6}.

∴ The number of favourable outcomes to the event E₂ = 1.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{1}{6}.$

Hence, the probability of getting a number greater than 5 is $\frac{1}{6}$.

(iv) Let E₃ be the event of getting a prime number, then

E₃ = {2, 3, 5}.

∴ The number of favourable outcomes to the event E₃ = 3.

$\therefore P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{3}{6} = \frac{1}{2}.$

Hence, the probability of getting a prime number is $\frac{1}{2}$.

(v) Let E₄ be the event of getting a number less than 7, then

E₄ = {1, 2, 3, 4, 5, 6}.

∴ The number of favourable outcomes to the event E₄ = 6.

$\therefore P(E_4) = \frac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \frac{6}{6} = 1.$

Hence, the probability of getting a number less than 7 is 1.

(vi) Let E₅ be the event of getting a number divisible by 3, then

E₅ = {3, 6}.

∴ The number of favourable outcomes to the event E₅ = 2.

$\therefore P(E_5) = \frac{\text{No. of favourable outcomes to } E_5}{\text{Total no. of possible outcomes}} = \frac{2}{6} = \frac{1}{3}.$

Hence, the probability of getting a number divisible by 3 is $\frac{1}{3}$.

(vii) Let E₆ be the event of getting a number between 3 and 6.

E₆ = {4, 5}.

∴ The number of favourable outcomes to the event E₆ = 2.

$\therefore P(E_6) = \frac{\text{No. of favourable outcomes to } E_6}{\text{Total no. of possible outcomes}} = \frac{2}{6} = \frac{1}{3}.$

Hence, the probability of getting a number between 3 and 6 is $\frac{1}{3}$.

(viii) Let E₇ be the event of getting a number divisible by 2 or 3, then

E₇ = {2, 3, 4, 6}.

∴ The number of favourable outcomes to the event E₇ = 4.

$\therefore P(E_7) = \frac{\text{No. of favourable outcomes to } E_7}{\text{Total no. of possible outcomes}} = \frac{4}{6} = \frac{2}{3}.$

Hence, the probability of getting a number divisible by 2 or 3 is $\frac{2}{3}$.

On a single throw of die,

Sample space = {1, 2, 3, -1, -2, -3}.

(i) Let E₁ be the event of getting a positive integer, then

E₁ = {1, 2, 3}.

∴ The number of favourable outcomes to the event E₁ = 3.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{3}{6} = \frac{1}{2}.$

Hence, the probability of getting a positive integer is $\frac{1}{2}$.

(ii) Let E₂ be the event of getting a integer greater than -3, then

E₂ = {-2, -1, 1, 2, 3}.

∴ The number of favourable outcomes to the event E₂ = 5.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{5}{6}.$

Hence, the probability of getting a integer greater than -3 is $\frac{5}{6}$.

(iii) Let E₃ be the event of getting smallest integer, then

E₃ = {-3}.

∴ The number of favourable outcomes to the event E₃ = 1.

$\therefore P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{1}{6}.$

Hence, the probability of getting smallest integer is $\frac{1}{6}$.

Sample space = {1, 2, 3, 4, 5, 6, 7, 8}.

(i) Let E₁ be the event of getting 8, then

E₁ = {8}.

∴ The number of favourable outcomes to the event E₁ = 1.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{1}{8}.$

Hence, the probability of getting 8 is $\frac{1}{8}$.

(ii) Let E₂ be the event of getting an odd number, then

E₂ = {1, 3, 5, 7}.

∴ The number of favourable outcomes to the event E₂ = 4.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{4}{8} = \frac{1}{2}.$

Hence, the probability of getting an odd number is $\frac{1}{2}$.

(iii) Let E₃ be the event of getting a number greater than 2, then

E₃ = {3, 4, 5, 6, 7, 8}.

∴ The number of favourable outcomes to the event E₃ = 6.

$\therefore P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{6}{8} = \frac{3}{4}.$

Hence, the probability of getting a number greater than 2 is $\frac{3}{4}$.

(iv) Let E₄ be the event of getting a number less than 9, then

E₄ = {1, 2, 3, 4, 5, 6, 7, 8}.

∴ The number of favourable outcomes to the event E₄ = 8.

$\therefore P(E_4) = \frac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \frac{8}{8} = 1.$

Hence, the probability of of getting a number less than 9 is 1.

In January there are 31 days and in an ordinary year there are 365 days but in a leap year there are 366 days.

(i) In January of a leap year, there are 31 days i.e., 4 weeks and 3 days. Therefore we have to find the probability of having a Monday out of the remaining 3 days.

Now 3 days can be (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday), (Wednesday, Thursday, Friday), (Thursday, Friday, Saturday), (Friday, Saturday, Sunday), (Saturday, Sunday, Monday), (Sunday, Monday, Tuesday).

In above 7 pairs, 3 times Monday occurs,

∴ Probability(having 5 Mondays) = $\frac{3}{7}$.

Hence, the probability that the month of January may have 5 Mondays in a leap year is $\frac{3}{7}$.

(ii) In January of an ordinary year, there are 31 days i.e. 4 weeks and 3 days. Out of remaining 3 days any one can be a monday,

Now 3 days can be (Monday, Tuesday, Wednesday), (Tuesday, Wednesday, Thursday), (Wednesday, Thursday, Friday), (Thursday, Friday, Saturday), (Friday, Saturday, Sunday), (Saturday, Sunday, Monday), (Sunday, Monday, Tuesday).

In above 7 pairs, 3 times Monday occurs,

∴ Probability (having 5 Mondays) = $\frac{3}{7}$.

Hence, the probability that the month of January may have 5 Mondays in a non-leap year is $\frac{3}{7}$.

In the month of February there are 29 days in a leap year while 28 days in a non-leap year.

(i) In a leap year's February there are 29 days i.e. 4 weeks and 1 day. In order to have 5 wednesdays, the remaining 1 day should be a wednesday

∴ Probability (having 5 wednesdays) = $\frac{1}{7}$.

Hence, the probability that the month of February may have 5 Wednesdays in a leap year is $\frac{1}{7}$.

(ii) In a non-leap year's February there are 28 days i.e. 4 weeks and 0 day.

∴ Probability (having 5 wednesdays) = $\frac{0}{7}$ = 0.

Hence, the probability that the month of February may have 5 Wednesdays in a non-leap year is 0.

On drawing a card from the box,

Sample space = {a, b, c, ......, m, n, o, p}.

(i) Let E₁ be the event of drawing a vowel card.

E₁ = {a, e, i, o}

∴ The number of favourable outcomes to the event E₁ = 4.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{4}{16} = \frac{1}{4}.$

Hence, the probability of drawing a vowel card is $\frac{1}{4}$.

(ii) Let E₂ be the event of drawing a consonant card.

Since, there are 4 vowels in the range a, b, ....., p. Hence, no. of consonants = 16 - 4 = 12.

∴ The number of favourable outcomes to the event E₂ = 12.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{12}{16} = \frac{3}{4}.$

Hence, the probability of drawing a consonant card is $\frac{3}{4}$.

(iii) Let E₃ be the event of drawing a card not containing letters of word median.

Since, there are 6 letters in the word median. Hence, no. of other letters = 16 - 6 = 10.

∴ The number of favourable outcomes to the event E₃ = 10.

$\therefore P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{10}{16}= \frac{5}{8}.$

Hence, the probability of drawing a card not containing letters of word 'median' is $\frac{5}{8}$.

(i) No. of consonants in the word 'BOUNDARIES' = 5 [B, N, D, R, S]

∴ No. of favourable outcomes = 5

Total no. of letters in the word 'BOUNDARIES' = 10.

∴ No. of possible outcomes = 10

P(that card drawn has a consonant) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{5}{10} = \frac{1}{2}$

Hence, probability of getting a consonant = $\frac{1}{2}$.

(ii) No. of different letters in the word 'LUCKNOW' that are also present in the word 'BOUNDARIES'` = 3 (U, N, O)

∴ No. of favourable outcomes = 3

P(that the letter on card is a letter of the word 'LUCKNOW') = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{3}{10}$

Hence, probability of getting one of the letter of the word 'LUCKNOW' = $\frac{3}{10}$.

(iii) No. of different letters in the word 'INDIA' that are also present in the word 'BOUNDARIES'` = 4 (I, N, D, A)

∴ No. of favourable outcomes = 4

P(that the letter on card is a letter of the word 'INDIA') = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{4}{10} = \frac{2}{5}$

Hence, probability of getting one of the letter of the word 'INDIA' = $\frac{2}{5}$.

On choosing an integer between 0 and 100,

Sample space = {1, 2, 3, 4, ........, 99}.

(i) Let E₁ be the event of choosing a number that is divisible by 7.

E₁ = {7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}.

∴ The number of favourable outcomes to the event E₁ = 14.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{14}{99}.$

Hence, the probability of choosing an integer divisible by 7 is $\frac{14}{99}$.

(ii) Let E₂ be the event of choosing a number that is not divisible by 7.

Since there are 14 numbers between 0 and 100 that are divisible by 7, hence the numbers not divisible by 7 = 99 - 14 = 85.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{85}{99}.$

Hence, the probability of choosing an integer not divisible by 7 is $\frac{85}{99}.$

Cards are marked 1, 2, 3, 4, ......, 20 and a card is drawn at random.

Sample space = {1, 2, 3, 4, ......, 20}, which has 20 equally likely outcomes.

(i) Let E₁ be the event of choosing a prime number.

E₁ = {2, 3, 5, 7, 11, 13, 17, 19}.

∴ The number of favourable outcomes to the event E₁ = 8.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{8}{20} = \frac{2}{5}.$

Hence, the probability of choosing a prime number is $\frac{2}{5}$.

(ii) Let E₂ be the event of choosing a number that is divisible by 3.

E₂ = {3, 6, 9, 12, 15, 18}.

∴ The number of favourable outcomes to the event E₂ = 6.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{6}{20} = \frac{3}{10}.$

Hence, the probability of choosing a number divisible by 3 is $\frac{3}{10}$.

(iii) Let E₃ be the event of choosing a perfect square.

E₃ = {1, 4, 9, 16}.

∴ The number of favourable outcomes to the event E₃ = 4.

$\therefore P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{4}{20} = \frac{1}{5}.$

Hence, the probability of choosing a perfect square is $\frac{1}{5}$.

(i) Let E₁ be the event of choosing an odd number disc.

E₁ = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}.

∴ The number of favourable outcomes to the event E₁ = 13.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{13}{25}.$

Hence, the probability of choosing an odd number disc is $\frac{13}{25}$.

(ii) Let E₂ be the event of choosing a disc with number that is divisible by both 2 and 3.

E₂ = {6, 12, 18, 24}.

∴ The number of favourable outcomes to the event E₂ = 4.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{4}{25}.$

Hence, the probability of choosing a disc with number that is divisible by both 2 and 3 is $\frac{4}{25}$.

(iii) Let E₃ be the event of choosing a disc with number less than 16.

E₃ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}.

∴ The number of favourable outcomes to the event E₃ = 15.

$\therefore P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{15}{25} = \frac{3}{5}.$

Hence, the probability of choosing a disc with number less than 16 is $\frac{3}{5}$.

(i) Let E₁ be the event of choosing an odd number card.

E₁ = {1, 3, 5, 7, 9, 11, 13, 15}.

∴ The number of favourable outcomes to the event E₁ = 8.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{8}{15}.$

Hence, the probability of choosing an odd number card is $\frac{8}{15}$.

(ii) Let E₂ be the event of choosing a prime number card.

E₂ = {2, 3, 5, 7, 11, 13}.

∴ The number of favourable outcomes to the event E₂ = 6.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{6}{15} = \frac{2}{5}.$

Hence, the probability of choosing a prime number card is $\frac{2}{5}$.

(iii) Let E₃ be the event of choosing card with number that is divisible by 3.

E₃ = {3, 6, 9, 12, 15}.

∴ The number of favourable outcomes to the event E₃ = 5.

$\therefore P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{5}{15} = \frac{1}{3}.$

Hence, the probability of choosing a card with number that is divisible by 3 is $\frac{1}{3}$.

(iv) Let E₄ be the event of choosing card with number that is divisible by 3 and 2.

E₄ = {6, 12}.

∴ The number of favourable outcomes to the event E₄ = 2.

$\therefore P(E_4) = \frac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \frac{2}{15}.$

Hence, the probability of choosing a card with number that is divisible by 3 and 2 is $\frac{2}{15}$.

(v) Let E₅ be the event of choosing card with number that is divisible by 3 or 2.

E₅ = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15}.

∴ The number of favourable outcomes to the event E₅ = 10.

$\therefore P(E_5) = \frac{\text{No. of favourable outcomes to } E_5}{\text{Total no. of possible outcomes}} = \frac{10}{15} = \frac{2}{3}.$

Hence, the probability of choosing a card with number that is divisible by 3 or 2 is $\frac{2}{3}$.

(vi) Let E₆ be the event of choosing card with perfect square number.

E₆ = {1, 4, 9}.

∴ The number of favourable outcomes to the event E₆ = 3.

$\therefore P(E_6) = \frac{\text{No. of favourable outcomes to } E_6}{\text{Total no. of possible outcomes}} = \frac{3}{15} = \frac{1}{5}.$

Hence, the probability of choosing a card with perfect square number is $\frac{1}{5}$.

Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. Hence, total no. of cards = 10.

(i) Let E₁ be the event of choosing a prime number card.

E₁ = {2}.

∴ The number of favourable outcomes to the event E₁ = 1.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{1}{10}.$

Hence, the probability of choosing a prime number card is $\frac{1}{10}$.

(ii) Let E₂ be the event of choosing a card with number that is divisible by 4.

E₂ = {4, 8, 12, 16, 20}.

∴ The number of favourable outcomes to the event E₂ = 5.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{5}{10} = \frac{1}{2}.$

Hence, the probability of choosing a card with number that is divisible by 4 is $\frac{1}{2}$.

(iii) Let E₃ be the event of choosing card with number that is a multiple of 6.

E₃ = {6, 12, 18}.

∴ The number of favourable outcomes to the event E₃ = 3.

$\therefore P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{3}{10}.$

Hence, the probability of choosing a card with number that is multiple of 6 is $\frac{3}{10}$.

(iv) Let E₄ be the event of choosing card with odd number.

E₄ = {}.

∴ The number of favourable outcomes to the event E₄ = 0.

$\therefore P(E_4) = \frac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \frac{0}{10} = 0.$

Hence, the probability of choosing a card with odd number is 0.

The cards are mixed thoroughly and a card is drawn at random from the box means that all the outcomes are equally likely.

Sample space = {13, 14, 15, ...., 60}, which has 48 equally likely outcomes.

(i) Let E₁ be the event of choosing card with number that is divisible by 5.

E₁ = {15, 20, 25, 30, 35, 40, 45, 50, 55, 60}.

∴ The number of favourable outcomes to the event E₁ = 10.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{10}{48} = \frac{5}{24}.$

Hence, the probability of choosing a card with number that is divisible by 5 is $\frac{5}{24}$.

(ii) Let E₂ be the event of choosing card with perfect square number.

E₂ = {16, 25, 36, 49}.

∴ The number of favourable outcomes to the event E₂ = 4.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{4}{48} = \frac{1}{12}.$

Hence, the probability of choosing a card with perfect square number is $\frac{1}{12}$.

Tickets are mixed thoroughly and a ticket is drawn at random from the box means that all the outcomes are equally likely.

Sample space = {3, 5, 7, 9, ...., 29}, which has 14 equally likely outcomes.

(i) Let E₁ be the event of choosing ticket with prime number.

E₁ = {3, 5, 7, 11, 13, 17, 19, 23, 29}.

∴ The number of favourable outcomes to the event E₁ = 9.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{9}{14}.$

Hence, the probability of choosing a ticket with prime number is $\frac{9}{14}$.

(ii) Let E₂ be the event of choosing ticket with number less than 16.

E₂ = {3, 5, 7, 9, 11, 13, 15}.

∴ The number of favourable outcomes to the event E₂ = 7.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{7}{14} = \frac{1}{2}.$

Hence, the probability of choosing a ticket with number less than 16 is $\frac{1}{2}$.

(iii) Let E₃ be the event of choosing ticket with number divisible by 3.

E₃ = {3, 9, 15, 21, 27}.

∴ The number of favourable outcomes to the event E₃ = 5.

$\therefore P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{5}{14}.$

Hence, the probability of choosing a ticket with number divisible by 3 is $\frac{5}{14}$.

A disc is drawn at random from the box means that all the outcomes are equally likely.

Sample space = {1, 2, 3, ....., 90}, which has 90 equally likely outcomes.

(i) Let E₁ be the event of drawing a disc with two digit number.

E₁ = {10, 11, 12, 13, ......, 90}.

∴ The number of favourable outcomes to the event E₁ = 81.

$\therefore P(E_1) = \frac{\text{No. of favourable outcomes to } E_1}{\text{Total no. of possible outcomes}} = \frac{81}{90} = \frac{9}{10}.$

Hence, the probability of drawing a disc with two digit number is $\frac{9}{10}.$

(ii) Let E₂ be the event of drawing a disc with perfect square number.

E₂ = {1, 4, 9, 16, 25, 36, 49, 64, 81}.

∴ The number of favourable outcomes to the event E₂ = 9.

$\therefore P(E_2) = \frac{\text{No. of favourable outcomes to } E_2}{\text{Total no. of possible outcomes}} = \frac{9}{90} = \frac{1}{10}.$

Hence, the probability of drawing a disc with perfect square number is $\frac{1}{10}$.

(iii) Let E₃ be the event of drawing a disc with number divisible by 5.

E₃ = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}.

∴ The number of favourable outcomes to the event E₃ = 18.

$\therefore P(E_3) = \frac{\text{No. of favourable outcomes to } E_3}{\text{Total no. of possible outcomes}} = \frac{18}{90} = \frac{1}{5}.$

Hence, the probability of drawing a disc with number that is divisible by 5 is $\frac{1}{5}$.

(iv) Let E₄ be the event of drawing a disc with prime number less than 30.

E₄ = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}.

∴ The number of favourable outcomes to the event E₄ = 10.

$\therefore P(E_4) = \frac{\text{No. of favourable outcomes to } E_4}{\text{Total no. of possible outcomes}} = \frac{10}{90} = \frac{1}{9}.$

Hence, the probability of drawing a disc with prime number less than 30 is $\frac{1}{9}$.

Let no. of white balls in bag = x, so no. of red balls = 15 - x.

P(drawing a white ball) = $\frac{\text{No. of favourable outcomes}}{\text{Total outcomes}} = \frac{x}{15}$

P(drawing a red ball) = $\frac{\text{No. of favourable outcomes}}{\text{Total outcomes}} = \frac{15 - x}{15}$.

Given, P(drawing a red ball) = 2 × P(drawing a white ball)

$$\begin{aligned}\therefore \frac{15 - x}{15} = \frac{2x}{15} \\[1em] \Rightarrow 15 - x = 2x \\[1em] \Rightarrow 2x + x = 15 \\[1em] \Rightarrow 3x = 15 \\[1em] \Rightarrow x = 5.\end{aligned}$$

Hence, there are 5 white balls in the bag.

Let no. of blue balls be x, total no. of balls = x + 6.

P(drawing a blue ball) = $\frac{\text{No. of favourable outcomes}}{\text{Total outcomes}} = \frac{x}{x + 6}$

P(drawing a red ball) = $\frac{\text{No. of favourable outcomes}}{\text{Total outcomes}} = \frac{6}{x + 6}$.

Given, P(drawing a blue ball) = 2 × P(drawing a red ball)

$$\begin{aligned}\therefore \frac{x}{x + 6} = 2 \times \frac{6}{x + 6} \\[1em] \Rightarrow \frac{x}{x + 6} = \frac{12}{x + 6} \\[1em] \Rightarrow x = 12.\end{aligned}$$

Total no. of balls = x + 6 = 12 + 6 = 18.

Hence, there are 18 balls in the bag.

Total balls = x + 2x + 3x = 24.

⇒ 6x = 24
⇒ x = 4.

(i) P(drawing a white ball) = $\frac{\text{No. of white balls}}{\text{Total balls}} = \frac{2 \times 4}{24} = \frac{1}{3}.$

Hence, the probability of drawing a white ball is $\frac{1}{3}$.

(ii) Total no. of white and blue balls = 2x + 3x = 5x = 20.

P(A) = P(drawing a not red ball) = P(drawing a white or blue ball).

$$\begin{aligned}\therefore P(A) = \frac{\text{Total no. of white and blue balls}}{\text{Total balls}} \\[1em] = \frac{20}{24} \\[1em] = \frac{5}{6}.\end{aligned}$$

Hence, the probability of not drawing a red ball is $\frac{5}{6}$.

Well-shuffling ensures equally likely outcomes.

Total number of outcomes = 52.

(i) There is only one 2 of spades in the whole pack.

∴ P('2' of spades) = $\frac{1}{52}$.

Hence, the probability of drawing 2 of spades = $\frac{1}{52}$.

(ii) There are 4 jacks, one of each suit.

∴ The number of favourable outcomes to the event 'a jack' = 4.

∴ P(a jack) = $\frac{4}{52} = \frac{1}{13}$.

Hence, the probability of drawing a jack = $\frac{1}{13}$.

(iii) There are two king of red colour, one of hearts and one of diamond.

∴ The number of favourable outcomes to the event 'king of red colour' = 2.

∴ P(king of red colour) = $\frac{2}{52} = \frac{1}{26}.$

Hence, the probability of drawing a king of red colour = $\frac{1}{26}$.

(iv) There are 13 cards of diamond suit.

∴ The number of favourable outcomes to the event 'a card of diamond' = 13.

∴ P(a card of diamond) = $\frac{13}{52} = \frac{1}{4}$.

Hence, the probability of drawing a diamond card = $\frac{1}{4}$.

(v) There are 8 king and queen cards, 2 in each suit.

∴ The number of favourable outcomes to the event 'a king or queen' = 8.

∴ P(a king or queen) = $\frac{8}{52} = \frac{2}{13}$.

Hence, the probability of drawing a king or queen = $\frac{2}{13}$.

(vi) There are 12 face cards.

∴ No. of non-face cards = 52 - 12 = 40.

∴ The number of favourable outcomes to the event 'a non-face card' = 40.

∴ P(a non-face card) = $\frac{40}{52} = \frac{10}{13}$.

Hence, the probability of drawing a non-face card = $\frac{10}{13}$.

(vii) Since 2 suits are of black colour and each suit has 3 face cards.

∴ No. of black face cards = 2 × 3 = 6.

∴ The number of favourable outcomes to the event 'a black face card' = 6.

∴ P(a black face card) = $\frac{6}{52} = \frac{3}{26}$.

Hence, the probability of drawing a black face card = $\frac{3}{26}$.

(viii) There are 2 suits of black cards.

∴ No. of black cards = 26.

∴ The number of favourable outcomes to the event 'a black card' = 26.

∴ P(a black card) = $\frac{26}{52} = \frac{1}{2}$.

Hence, the probability of drawing a black card = $\frac{1}{2}$.

(ix) There are 4 ace cards, one in each suit.

∴ No. of non-ace cards = 52 - 4 = 48.

∴ The number of favourable outcomes to the event 'a non-ace card' = 48.

∴ P(a non-ace card) = $\frac{48}{52} = \frac{12}{13}$.

Hence, the probability of drawing a non-ace card = $\frac{12}{13}$.

(x) There are 3 face cards in each suit and 2 suits of black colour.

Hence, no. of face cards of black colour = 6.

∴ No. of non-face black cards = 26 - 6 = 20.

∴ The number of favourable outcomes to the event 'a non-face card of black colour' = 20.

∴ P(a non-face black card) = $\frac{20}{52} = \frac{5}{13}$.

Hence, the probability of drawing a non-face black card = $\frac{5}{13}$.

(xi) There are 13 spade cards and each suit has 1 jack.

So, the other 3 suits apart from spade has 3 jacks.

∴ Total no. of spade and jack cards = 13 + 3 = 16.

Hence, no. of cards other than spade and jack = 52 - 16 = 36.

∴ The number of favourable outcomes to the event 'neither a spade nor a jack' = 36.

∴ P(neither a spade nor a jack) = $\frac{36}{52} = \frac{9}{13}$.

Hence, the probability of drawing neither a spade nor a jack = $\frac{9}{13}$.

(xii) There are 13 heart cards and 2 suits of red colour.

Since, one king of red colour is already included in hearts hence only one red king is more.

∴ Total no. of heart and red king cards = 13 + 1 = 14.

Hence, no. of cards other than heart and red king cards = 52 - 14 = 38.

∴ The number of favourable outcomes to the event 'neither heart nor a red king' = 38.

∴ P(neither a heart nor a red king) = $\frac{38}{52} = \frac{19}{26}$.

Hence, the probability of drawing neither a heart nor a red king = $\frac{19}{26}$.

3 face cards of spades are removed, hence total cards = 52 - 3 = 49.

Well-shuffling ensures equally likely outcomes.

Total number of outcomes = 49.

(i) Since 2 suits are of black colour and each suit has 3 face cards but since spades of black colour are removed.

∴ No. of black face cards = 3.

∴ The number of favourable outcomes to the event 'a black face card' = 3.

∴ P(a black face card) = $\frac{3}{49}$.

Hence, the probability of drawing a black face card = $\frac{3}{49}$.

(ii) Each suit has one queen.

Since, face cards of spades are removed hence, it has no queen.

So, there are 3 queens left.

∴ The number of favourable outcomes to the event 'a queen' = 3.

∴ P(a queen) = $\frac{3}{49}$.

Hence, the probability of drawing a queen = $\frac{3}{49}$.

(iii) Total no. of black cards = 26.

Since, spades are of black colour and it's face card are removed.

∴ The number of black cards left = 26 - 3 = 23.

∴ The number of favourable outcomes to the event 'a black card' = 23.

∴ P(a black card) = $\frac{23}{49}$.

Hence, the probability of drawing a black card = $\frac{23}{49}$.

(iv) There are 13 heart cards.

∴ The number of favourable outcomes to the event 'a heart' = 13.

∴ P(a heart) = $\frac{13}{49}$.

Hence, the probability of drawing a heart = $\frac{13}{49}$.

(v) Since, face cards of spades are removed,

∴ No. of spades left = 13 - 3 = 10.

∴ P(a spade) = $\frac{10}{49}$.

Hence, the probability of drawing a spade = $\frac{10}{49}$.

(vi) There are 2, '9' numbered cards of black colour.

∴ P(a '9' of black colour) = $\frac{2}{49}$.

Hence, the probability of drawing a '9' of black colour = $\frac{2}{49}$.

Given, a black jack, a red queen and two black kings fell down.

Hence, remaining no. of cards = 52 - 4 = 48.

(i) There are total 26 black cards, 13 of club and 13 of spades. Out of which 3 are removed.

∴ No. of black cards left = 26 - 3 = 23.

P(a black card) = $\frac{23}{48}.$

Hence, the probability of drawing a black card is $\frac{23}{48}.$

(ii) There are total 4 kings, but 2 black kings are removed.

∴ No. of kings left = 4 - 2 = 2.

P(a king) = $\frac{2}{48} = \frac{1}{24}.$

Hence, the probability of drawing a king is $\frac{1}{24}.$

(iii) There are 2 red queens, one of hearts and one of diamonds.

Since, one red queen is removed, hence no. of red queens left = 2 - 1 = 1.

P(a red queen) = $\frac{1}{48}.$

Hence, the probability of drawing a red queen is $\frac{1}{48}.$

When two different coins are tossed simultaneously, then sample space = {HH, HT, TH, TT}. It consist of 4 equally likely outcomes.

Total number of possible outcomes = 4.

(i) Let A be the event 'two heads', then A = {HH} and the number of outcomes favourable to the event A = 1.

∴ P(two heads) = $\frac{1}{4}$.

Hence, the probability of getting two heads = $\frac{1}{4}$.

(ii) Let B be the event 'atleast one tail', then B = {HT, TT, TH} and the number of outcomes favourable to the event B = 3.

∴ P(atleast one tail) = $\frac{3}{4}$.

Hence, the probability of getting atleast one tail = $\frac{3}{4}$.

When two different coins are tossed simultaneously, then sample space = {HH, HT, TH, TT}. It consists of 4 equally likely outcomes.

Total number of possible outcomes = 4.

(i) Let A be the event 'two tails', then A = {TT} and the number of outcomes favourable to the event A = 1.

∴ P(A) = $\frac{1}{4}$.

Hence, the probability of getting two tails = $\frac{1}{4}$.

(ii) Let B be the event 'one tail', then A = {HT, TH} and the number of outcomes favourable to the event B = 2.

∴ P(B) = $\frac{2}{4} = \frac{1}{2}$.

Hence, the probability of getting one tail = $\frac{1}{2}$.

(iii) Let C be the event 'no tail', then C = {HH} and the number of outcomes favourable to the event C = 1.

∴ P(C) = $\frac{1}{4}$.

Hence, the probability of getting no tail = $\frac{1}{4}$.

(iv) Let D be the event 'atmost one tail', then D = {HH, HT, TH} and the number of outcomes favourable to the event D = 3.

∴ P(D) = $\frac{3}{4}$.

Hence, the probability of getting atmost one tail = $\frac{3}{4}$.

When two different dice are rolled together, the total number of outcomes is 6 × 6 i.e. 36 and all outcomes are equally likely. The sample space of the random experiment has 36 equally likely outcomes. The sample of the experiment

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
       (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
       (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
       (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
       (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
       (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).}

It consists of 36 equally likely outcomes.

(i) Let A be the event of getting 'a number greater than 3 on each dice', then

A = {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}.

∴ The number of outcomes favourable to event A = 9.

∴ P(A) = $\frac{9}{36} = \frac{1}{4}.$

Hence, the probability of getting a number greater than 3 on each dice is $\frac{1}{4}$.

(ii) Let B be the event of getting 'an odd number on both dice', then

A = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}.

∴ The number of outcomes favourable to event B = 9.

∴ P(B) = $\frac{9}{36} = \frac{1}{4}.$

Hence, the probability of getting an odd number on both the dice is $\frac{1}{4}$.

(i) Let A be the event of getting 'a doublet', then

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}.

∴ The number of outcomes favourable to event A = 6.

∴ P(A) = $\frac{6}{36} = \frac{1}{6}.$

Hence, the probability of getting a doublet is $\frac{1}{6}$.

(ii) Let B be the event of getting 'a sum of 8', then

A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}.

∴ The number of outcomes favourable to event B = 5.

∴ P(B) = $\frac{5}{36}.$

Hence, the probability of getting a sum of 8 is $\frac{5}{36}$.

(iii) Let C be the event of getting 'a sum divisible by 5', then

C = {(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)}.

∴ The number of outcomes favourable to event C = 7.

∴ P(C) = $\frac{7}{36}.$

Hence, the probability of getting a sum divisible by 5 is $\frac{7}{36}$.

(iv) Let D be the event of getting 'sum of atleast 11', then

D = {(5, 6), (6, 5), (6, 6)}.

∴ The number of outcomes favourable to event D = 3.

∴ P(D) = $\frac{3}{36} = \frac{1}{12}.$

Hence, the probability of getting a sum of atleast 11 is $\frac{1}{12}$.

Letters written on cards = {'A', 'D', 'M', 'N', 'O', 'S', 'U', 'Y'}

No. of cards = 8

(a) Letters of the word MONDAY present in the cards = {'M', 'O', 'N', 'D', 'A', 'Y'}

Probability that the card drawn is a letter of the word MONDAY

= $\frac{\text{Letters of MONDAY present}}{\text{No. of cards}} = \frac{6}{8} = \frac{3}{4}$.

Hence, required probability = $\frac{3}{4}$.

(b) Letters of the word not present in MONDAY = {'S', 'U'}

Probability that the card drawn is not a letter of the word MONDAY

= $\frac{\text{Letters not present in MONDAY}}{\text{No. of cards}} = \frac{2}{8} = \frac{1}{4}$.

Hence, required probability = $\frac{1}{4}$.

(c) Letters of the word present in SUNDAY and MONDAY are {'N', 'D', 'A', 'Y'}.

Probability that the card drawn has a letter which appears both in SUNDAY and MONDAY

= $\frac{\text{Letters present in both MONDAY and SUNDAY}}{\text{No. of cards}} = \frac{4}{8} = \frac{1}{2}$.

Hence, required probability = $\frac{1}{2}$.

(a) Since, there are three people which can be called.

Hence, probability that Akbar is chosen for the call = $\frac{1}{3}$.

(b) Since, there are four options out of which one is correct.

Hence, probability that Akbar couldn't give the correct answer = $\frac{3}{4}$.

Since, the probability cannot be negative.

Hence, Option 3 is the correct option.

Probability of an event is p.

∴ Probability of complementary event is 1 - p.

Hence, Option 3 is the correct option.

Sample space = {2, 3, 5, 7}.

Let A be the event of selecting an even number,

∴ A = {2}.

Hence, the no. of favourable outcomes to A = 1.

∴ P(A) = $\frac{1}{4}.$

Hence, Option 2 is the correct option.

When a die is thrown the sample space is,

S = {1, 2, 3, 4, 5, 6}.

Let A be the event of getting an odd number less than 3,

∴ A = {1}.

Hence, the no. of favourable outcomes to A = 1.

∴ P(A) = $\frac{1}{6}.$

Hence, Option 1 is the correct option.

There are 2 numbers between 1 to 6 that are divisible by 3, i.e., 3 and 6.

∴ No. of favourable outcomes = 2

Total no. in a dice = 6.

∴ No. of possible outcomes = 6

P(a number divisible by 3) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{2}{6} = \frac{1}{3}$.

Hence, option 2 is the correct option.

When a die is thrown the sample space is,

S = {1, 2, 3, 4, 5, 6}.

Let A be the event of getting an even prime number,

∴ A = {2}.

Hence, the no. of favourable outcomes to A = 1.

∴ P(A) = $\frac{1}{6}.$

Hence, Option 1 is the correct option.

When a die is thrown the sample space is,

S = {1, 2, 3, 4, 5, 6}.

Let A be the event of getting a composite number,

∴ A = {4, 6}.

Hence, the no. of favourable outcomes to A = 2.

∴ P(A) = $\frac{2}{6} = \frac{1}{3}.$

Hence, Option 1 is the correct option.

When a die is thrown the sample space is,

S = {1, 2, 3, 4, 5, 6}.

Let A be the event of getting an even number or a number greater than 4,

∴ A = {2, 4, 5, 6}.

Hence, the no. of favourable outcomes to A = 4.

∴ P(A) = $\frac{4}{6} = \frac{2}{3}.$

Hence, Option 4 is the correct option.

Total no. of letters in English alphabet = 26.

Let A be the event of getting a letter of the word 'DELHI',

∴ A = {D, E, L, H, I}.

Hence, the no. of favourable outcomes to A = 5.

∴ P(A) = $\frac{5}{26}.$

Hence, Option 3 is the correct option.

There are 6 red face cards, 3 of hearts and 3 of diamonds.

∴ The number of favourable outcomes to event 'red face card' = 6.

∴ P(red face card) = $\frac{6}{52} = \frac{3}{26}$.

Hence, Option 1 is the correct option.

Well-shuffling ensures equally likely outcomes.

Total number of outcomes = 52.

There are 4 kings and 4 jacks i.e. total 8 kings and jacks.

∴ The number of favourable outcomes to event 'being a king or jack' is = 8.

∴ P(a king or jack) = $\frac{8}{52} = \frac{2}{13}.$

Hence, Option 3 is the correct option.

Number of days in a non-leap year = 365 i.e. 52 weeks and 1 day.

In order to have 53 sundays, the 1 day should be a sunday.

Let A be the event of the day to be a Sunday,

∴ P(A) = $\frac{1}{7}$.

Hence, Option 4 is the correct option.

Probability that a ball drawn from the bag at random will be neither red nor black is equal to probability of picking a white ball.

No. of white balls = 5, total balls = 3 + 5 + 7 = 15.

∴ P(neither red nor black ball) = $\frac{5}{15} = \frac{1}{3}$.

Hence, Option 2 is the correct option.

Total no. of balls = 4 + 5 = 9.

Let A be the event of drawing a red or green ball. As, the total no. of red and green balls = 9, so the number of favourable outcomes to event A = 9.

∴ P(A) = $\frac{9}{9}$ = 1.

Hence, Option 4 is the correct option.

Sample space = {1, 2, 3, ......., 40}, with 40 equally likely outcomes.

Let A be the event that selected ticket has a number which is a multiple of 5.

∴ A = {5, 10, 15, 20, 25, 30, 35, 40}.

Hence, the number of favourable outcomes to event A = 8.

∴ P(A) = $\frac{8}{40} = \frac{1}{5}$.

Hence, Option 1 is the correct option.

Sample space = {1, 2, 3, ......., 25}, with 25 equally likely outcomes.

Let A be the event that chosen number is prime.

∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23}.

Hence, the number of favourable outcomes to event A = 9.

∴ P(A) = $\frac{9}{25}$.

Hence, Option 2 is the correct option.

Sample space = {1, 2, 3, ...., 90}, which has 90 equally likely outcomes.

Let A be the event of drawing a perfect square card.

∴ A = {1, 4, 9, 16, 25, 36, 49, 64, 81}.

Hence, the number of favourable outcomes to event A = 9.

∴ P(A) = $\frac{9}{90} = \frac{1}{10}$.

Hence, Option 1 is the correct option.

A coin is tossed twice,

Sample space = {HH, HT, TH, TT}.

Let A be the event of getting two heads,

∴ A = {HH}.

∴ P(A) = $\frac{1}{4}$.

Hence, Option 1 is the correct option.

When two different coins are tossed simultaneously, then sample space = {HH, HT, TH, TT}. It consist of 4 equally likely outcomes.

Total number of possible outcomes = 4.

Let A be the event of getting atleast one head,

∴ A = {HH, HT, TH}.

Hence, the no. of favourable outcomes to event A = 3.

∴ P(A) = $\frac{3}{4}$.

Hence, Option 3 is the correct option.

When two different coins are tossed simultaneously, then sample space = {HH, HT, TH, TT}. It consist of 4 equally likely outcomes.

Total number of possible outcomes = 4.

Let A be the event of getting atmost one head,

∴ A = {TT, HT, TH}.

Hence, the no. of favourable outcomes to event A = 3.

∴ P(A) = $\frac{3}{4}$.

Hence, Option 2 is the correct option.

Let the no. of bad eggs be x.

$$\begin{aligned}\therefore \text{ P(getting a bad egg)} = \frac{x}{400} \\[1em] \Rightarrow 0.035 = \frac{x}{400} \\[1em] \Rightarrow x = 0.035 \times 400 \\[1em] \Rightarrow x = 14.\end{aligned}$$

Hence, Option 2 is the correct option.

The sun always rises from east and also in a leap year February has 29 days.

∴ Event A and Event C has probability equal to 1.

Hence, Option 4 is the correct option.

Since, there is no black marble in the bag.

∴ Probability of drawing a black marble = 0.

Hence, Option 1 is the correct option.

In a leap year, there are 366 days.

366 days = 52 weeks + 2 days

These 2 days can be (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), and (Sun, Mon).

Total number of possible outcomes = 7

Number of favourable outcomes (Getting Sunday as one of the extra days) = 2 (i.e., (Sat, Sun), (Sun, Mon)).

P(Getting Sunday as one of the extra days) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{2}{7}$

∴ Assertion (A) is true.

In a non - leap year, there are 365 days.

365 days = 52 weeks + 1 days

These 1 days can be Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.

Total number of possible outcomes = 7

Number of favourable outcomes (Getting Sunday as one of the extra days) = 1

P(Getting Sunday as one of the extra days) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{1}{7}$

∴ Reason (R) is false.

∴ Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

Given, the probability of Sania winning the match = 0.79.

As we know that the sum of probabilities of two complementary event is 1.

⇒ P(Sania wins) + P(Ashma wins) = 1

∴ Reason (R) is true.

⇒ 0.79 + P(Ashma wins) = 1

⇒ P(Ashma wins) = 1 - P(Sania wins)

⇒ P(Ashma wins) = 1 - 0.79

⇒ P(Ashma wins) = 0.21

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given, the probability of occurrence of an event E = $\frac{5}{11}$

As we know that the sum of probabilities of two complementary event is 1.

⇒ P(E) + P($\overline{E}$) = 1

∴ Reason (R) is true.

$$\begin{aligned}\Rightarrow \frac{5}{11} + P(\overline{E}) = 1\\[1em] \Rightarrow P(\overline{E}) = 1 - \frac{5}{11}\\[1em] \Rightarrow P(\overline{E}) = \frac{11 - 5}{11}\\[1em] \Rightarrow P(\overline{E}) = \frac{6}{11}\end{aligned}$$

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

When two coins are tossed together, the total number of possible outcomes = 4 (i.e. HH, HT, TH and TT)

∴ Reason (R) is true.

Number of favourable outcomes (Getting one head) = 2 (HT and TH)

P(getting one head) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{2}{4} = \frac{1}{2}$

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

On the faces of a die, numbers are 1, 2, 3, 4, 5 and 6. Out of which 2, 3 and 5 are only prime numbers.

∴ Reason (R) is true.

Number of favourable outcomes (of getting prime number) = 3

Number of possible outcomes = 6

P(getting a prime number) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{3}{6} = \frac{1}{2}$

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

On the faces of a die, numbers are 1, 2, 3, 4, 5 and 6. Out of which 2, 4 and 6 are even numbers.

So, the sample space for even number on a die = {2, 4, 6}

∴ Reason (R) is true.

Number of favourable outcomes = 3 (for getting even numbers)

The total number of possible outcomes = 6

P(getting a even number) = $\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \frac{3}{6} = \frac{1}{2}$.

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

When a die is thrown once, the possible outcomes are the numbers 1, 2, 3, 4, 5, 6. So, the sample space of the experiment = {1, 2, 3, 4, 5, 6}. It has six equally likely outcomes.

(i) The event is getting number greater than 5 i.e. {6}

The number of favourable outcomes to the event getting number greater than 5 = 1.

∴ P(getting number greater than 5) = $\frac{1}{6}$.

Hence, the probability of getting a number greater than 5 is $\frac{1}{6}$.

(ii) The event is getting an odd prime number i.e. {3, 5}

The number of favourable outcomes to the event an odd prime number = 2.

∴ P(getting an odd prime number) = $\frac{2}{6} = \frac{1}{3}$.

Hence, the probability of getting an odd prime number is $\frac{1}{3}$.

(iii) The event is getting a number which is a multiple of 3 or 4 i.e. {3, 4, 6}

The number of favourable outcomes to the above event = 3.

∴ P(getting a number which is a multiple of 3 or 4) = $\frac{3}{6} = \frac{1}{2}$.

Hence, the probability of getting a number which is a multiple of 3 or 4 is $\frac{1}{2}$.

Total phones = 48.

(i) Varnika accepts good phones only, hence no. of favourable outcomes acceptable to Varnika = 42.

P(acceptable to Varnika) = $\frac{42}{48} = \frac{7}{8}.$

Hence, the probability that a phone is acceptable to Varnika is $\frac{7}{8}.$

(ii) Trader accepts good phones and one with minor defects only, hence no. of favourable outcomes acceptable to trader = 42 + 3 = 45.

P(acceptable to trader) = $\frac{45}{48} = \frac{15}{16}.$

Hence, the probability that a phone is acceptable to trader is $\frac{15}{16}.$

Since, a ball is drawn at random from the bag, so all the balls are equally likely to be drawn.

Total number of balls in the bag = 5 + 8 + 7 = 20.

So, the sample space of the experiment has 20 equally likely outcomes.

(i) Let E₁ be the event 'a red or white ball is drawn'.

The number of red or white balls = 5 + 8 = 13.

So, the number of favourable outcomes to the E₁ = 13.

∴ P(E₁) = P(a red or white ball) = $\frac{13}{20}.$

Hence, the probability that the ball drawn is red or white is $\frac{13}{20}.$

(ii) Drawing a not black ball means drawing a red or white ball.

∴ P(not black ball) = P(a red or white ball) = $\frac{13}{20}.$

Hence, the probability that the ball drawn is not a black ball is $\frac{13}{20}.$

(iii) If a ball drawn is neither white nor black it means a red ball is drawn.

No. of red balls = 5.

∴ P(a red ball is drawn) = $\frac{5}{20} = \frac{1}{4}.$

Hence, the probability that a red ball is drawn is $\frac{1}{4}.$

Since, a ball is drawn at random from the bag, so all the balls are equally likely to be drawn.

Total number of balls in the bag = 5 + 7 + 4 + 2 = 18.

So, the sample space of the experiment has 18 equally likely outcomes.

(i) Let E₁ be the event 'a white or blue ball is drawn'.

The number of white or blue balls = 5 + 2 = 7.

So, the number of favourable outcomes to the E₁ = 7.

∴ P(E₁) = P(a white or blue ball) = $\frac{7}{18}.$

Hence, the probability that the ball drawn is white or blue is $\frac{7}{18}.$

(ii) Let E₂ be the event 'a red or black ball is drawn'.

The number of red or black balls = 7 + 4 = 11.

So, the number of favourable outcomes to the E₂ = 11.

∴ P(E₂) = P(a red or black ball) = $\frac{11}{18}.$

Hence, the probability that the ball drawn is red or black is $\frac{11}{18}.$

(iii) Let E₃ be the event 'not white ball is drawn'.

Probability of not white ball drawn = Probability of any other ball drawn.

No. of balls except white balls = 18 - 5 = 13.

So, the number of favourable outcomes to the E₃ = 13.

∴ P(E₃) = P(not a white ball) = $\frac{13}{18}.$

Hence, the probability that the ball drawn is not white is $\frac{13}{18}.$

(iv) Let E₄ be the event 'neither white nor black ball is drawn'.

Probability that neither white nor black ball is drawn = Probability of a red or blue ball drawn.

No. of red or blue balls = 7 + 2 = 9.

So, the number of favourable outcomes to the E₄ = 9.

∴ P(E₄) = $\frac{9}{18} = \frac{1}{2}.$

Hence, the probability that the ball drawn is neither white nor black is $\frac{1}{2}.$

A ball is drawn at random from the box it means that all outcomes are equally likely.

Sample space = {1, 2, 3, ......, 20}, which has 20 equally likely outcomes.

(i) Let A be the event 'the number on the ball is odd', then

A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}.

∴ The number of favourable outcomes to the event A = 10.

∴ P(odd number) = $\frac{10}{20} = \frac{1}{2}.$

Hence, the probability that the ball drawn has an odd number is $\frac{1}{2}.$

(ii) Let B be the event 'the number on the ball is divisible by 2 or 3', then

B = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20}.

∴ The number of favourable outcomes to the event B = 13.

∴ P(number is divisible by 2 or 3) = $\frac{13}{20}.$

Hence, the probability that the ball drawn has a number that is divisible by 2 or 3 is $\frac{13}{20}.$

(iii) Let C be the event 'the number on the ball is prime', then

C = {2, 3, 5, 7, 11, 13, 17, 19}.

∴ The number of favourable outcomes to the event C = 8.

∴ P(prime number) = $\frac{8}{20} = \frac{2}{5}.$

Hence, the probability that the ball drawn has a prime number is $\frac{2}{5}.$

(iv) Let D be the event 'the number on the ball is not divisible by 10', then

D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19}.

∴ The number of favourable outcomes to the event D = 18.

∴ P(number is not divisible by 10) = $\frac{18}{20} = \frac{9}{10}.$

Hence, the probability that the ball drawn has a number that is not divisible by 10 is $\frac{9}{10}.$

A number is selected at random it means that all outcomes are equally likely.

Sample space = {1, 2, 3, ......, 35}, which has 35 equally likely outcomes.

(i) Let A be the event 'the number is prime', then

A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}.

∴ The number of favourable outcomes to the event A = 11.

∴ P(prime number) = $\frac{11}{35}.$

Hence, the probability that the number selected is a prime number is $\frac{11}{35}.$

(ii) Let B be the event 'the number is multiple of 7', then

B = {7, 14, 21, 28, 35}.

∴ The number of favourable outcomes to the event B = 5.

∴ P(multiple of 7) = $\frac{5}{35} = \frac{1}{7}.$

Hence, the probability that the number selected is a multiple of 7 is $\frac{1}{7}.$

(iii) Let C be the event 'the number is multiple of 3 or 5', then

C = {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35}.

∴ The number of favourable outcomes to the event C = 16.

∴ P(multiple of 3 or 5) = $\frac{16}{35}.$

Hence, the probability that the number selected is a multiple of 3 or 5 is $\frac{16}{35}.$

A card is selected at random it means that all outcomes are equally likely.

Sample space = {13, 14, 15, ......, 60}, which has 48 equally likely outcomes.

(i) Let A be the event 'the number is divisible by 5', then

A = {15, 20, 25, 30, 35, 40, 45, 50, 55, 60}.

∴ The number of favourable outcomes to the event A = 10.

∴ P(divisible by 5) = $\frac{10}{48} = \frac{5}{24}.$

Hence, the probability that the number selected is divisible by 5 is $\frac{5}{24}.$

(ii) Let B be the event 'the number is a perfect square', then

B = {16, 25, 36, 49}.

∴ The number of favourable outcomes to the event B = 4.

∴ P(perfect square) = $\frac{4}{48} = \frac{1}{12}.$

Hence, the probability that the number selected is a perfect square is $\frac{1}{12}.$

A card is selected at random it means that all outcomes are equally likely.

Sample space = {14, 15, 16, ......, 99}, which has 86 equally likely outcomes.

(i) Let A be the event 'odd number', then

A = {15, 17, 19, 21, 23, ........., 91, 93, 95, 97, 99}.

∴ The number of favourable outcomes to the event A = 43.

∴ P(odd number) = $\frac{43}{86} = \frac{1}{2}.$

Hence, the probability that the number selected is an odd number is $\frac{1}{2}.$

(ii) Let B be the event 'a perfect square number', then

B = {16, 25, 36, 49, 64, 81}.

∴ The number of favourable outcomes to the event B = 6.

∴ P(a perfect square number) = $\frac{6}{86} = \frac{3}{43}.$

Hence, the probability that the number selected is a perfect square number is $\frac{3}{43}.$

Let no. of blue balls be x, so total no. of balls = x + 5.

Given,

P(a blue ball drawn) = 4 × P(a red ball drawn) .

$$\begin{aligned}\Rightarrow \frac{x}{x + 5} = 4 \times \frac{5}{x + 5} \\[1em] \Rightarrow \frac{x}{x + 5} = \frac{20}{x + 5} \\[1em] \Rightarrow x = 20.\end{aligned}$$

Total no. of balls = x + 5 = 20 + 5 = 25.

(i) Let A be the event 'a white ball is drawn'.

∴ The no. of favourable outcomes to event A = x.

∴ P(a white ball is drawn) = $\frac{x}{18}.$

Hence, the probability that a white ball is drawn = $\frac{x}{18}$.

(ii) If 2 more white balls are added,

No. of white balls = x + 2,

Total no. of balls = 18 + 2 = 20.

Hence now,

P(a white ball is drawn) = $\frac{x + 2}{20}$

Given, probability of drawing a white ball will be $\frac{9}{8}$ times that of probability of white ball coming in part (i)

$$\begin{aligned}\therefore \frac{x + 2}{20} = \frac{9}{8} \times \frac{x}{18} \\[1em] \Rightarrow \frac{x + 2}{20} = \frac{x}{16} \\[1em] \Rightarrow 16(x + 2) = 20x \\[1em] \Rightarrow 16x + 32 = 20x \\[1em] \Rightarrow 20x - 16x = 32 \\[1em] \Rightarrow 4x = 32 \\[1em] \Rightarrow x = 8.\end{aligned}$$

Hence, the value of x = 8.

Well shuffling ensures equally likely outcomes.

Total number of outcomes = 52.

(i) Each suit has one king, one queen and one jack, and there are two suits of red colour.

∴ There are 2 kings, 2 queens and 2 jacks.

∴ The number of red face cards = 6.

∴ P(a red face card) = $\frac{6}{52} = \frac{3}{26}.$

Hence, the probability that a card drawn is a red face card = $\frac{3}{26}$.

(ii) There are 13 clubs and 13 spades i.e. total = 26 cards.

No. of cards left other than spades and clubs = 52 - 26 = 26.

P(neither a club nor spade) = $\frac{26}{52} = \frac{1}{2}.$

Hence, the probability that a card drawn is neither a club nor spade card = $\frac{1}{2}$.

(iii) There are 2 kings of red colour, one of heart and one of diamond.

There are 4 aces, one of each suit.

No. of cards other than ace and king of red colour = 52 - 4 - 2 = 46.

P(neither an ace nor a king of red colour) = $\frac{46}{52} = \frac{23}{26}.$

Hence, the probability that a card drawn is neither an ace nor a king of red colour = $\frac{23}{26}$.

(iv) There are 26 red cards, 13 of hearts and 13 of diamonds.

There are 4 queens, one of each suit but since red queens are included in red cards hence, queens left = 2.

Total no. of red cards and queen = 26 + 2 = 28.

No. of cards other than queen and red cards = 52 - 28 = 24.

P(neither a red card nor queen) = $\frac{24}{52} = \frac{6}{13}.$

Hence, the probability that a card drawn is neither a red card nor queen = $\frac{6}{13}$

(v) There are 26 red cards, 13 of hearts and 13 of diamonds.

There are 2 black kings, one of club and one of spade.

Total no. of red cards and black kings = 26 + 2 = 28.

No. of cards other than red cards and black kings = 52 - 28 = 24.

P(neither a red card nor black king) = $\frac{24}{52} = \frac{6}{13}.$

Hence, the probability that a card drawn is neither a red card nor black king = $\frac{6}{13}$

There are 2 black jacks, 2 black kings and 2 black aces.

No. of cards left after removing black jacks, black kings and black aces = 52 - 6 = 46.

(i) There are 26 red cards, 13 of hearts and 13 of diamonds.

∴ P(a red card) = $\frac{26}{46} = \frac{13}{23}.$

Hence, the probability of drawing a red card = $\frac{13}{23}.$

(ii) There are 12 face cards, out of which 4 are removed.

No. of face cards left = 8,

P(a face card) = $\frac{8}{46} = \frac{4}{23}.$

Hence, the probability of drawing a face card = $\frac{4}{23}.$

(iii) There are 13 cards in diamond and 13 cards in club.

Out of club, one king, one queen and one ace is removed,

No. of club cards left = 13 - 3 = 10.

Total no. of diamond and club cards = 13 + 10 = 23.

P(a diamond or club) = $\frac{23}{46} = \frac{1}{2}.$

Hence, the probability of drawing a diamond or club card = $\frac{1}{2}.$

(iv) There are 13 cards in spade and 4 queens (including the queen of spade).

Out of spade, one king, one queen and one ace is removed,

No. of spade cards left = 13 - 3 = 10.

3 queens are left after removing the queen of spade, as it is already included while counting spade cards.

Total no. of spade and queens = 10 + 3 = 13.

P(a spade or queen) = $\frac{13}{46}.$

Hence, the probability of drawing a spade or queen = $\frac{13}{46}.$

When two different dice are rolled together, the total number of outcomes is 6 × 6 i.e. 36 and all outcomes are equally likely. The sample space of the random experiment has 36 equally likely outcomes. The sample of the experiment

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
       (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
       (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
       (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
       (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
       (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).}

It consists of 36 equally likely outcomes.

(i) Let A be the event of getting a sum of 7,

A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

∴ The number of outcomes favourable to event A = 6.

∴ P(sum of 7) = $\frac{6}{36} = \frac{1}{6}.$

Hence, the probability of getting a sum of 7 is $\frac{1}{6}$.

(ii) Let B be the event of getting a sum ≤ 3,

B = {(1, 1), (1, 2), (2, 1)}

∴ The number of outcomes favourable to event B = 3.

∴ P(sum ≤ 3) = $\frac{3}{36} = \frac{1}{12}.$

Hence, the probability of getting a sum ≤ 3 is $\frac{1}{12}$.

(iii) Let C be the event of getting a sum ≤ 10,

C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (6, 1), (6, 2), (6, 3), (6, 4)}

∴ The number of outcomes favourable to event C = 33.

∴ P(sum ≤ 10) = $\frac{33}{36} = \frac{11}{12}.$

Hence, the probability of getting a sum ≤ 10 is $\frac{11}{12}$.

When two different dice are rolled together, the total number of outcomes is 6 × 6 i.e. 36 and all outcomes are equally likely. The sample space of the random experiment has 36 equally likely outcomes. The sample of the experiment

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
       (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
       (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
       (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
       (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
       (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).}

It consists of 36 equally likely outcomes.

(i) Let A be the event of getting numbers on top of dice with product 4.

A = {(1, 4), (2, 2), (4, 1)}.

∴ P(A) = $\frac{3}{36} = \frac{1}{12}.$

Hence, the probability that the product of the numbers on the top of two dice is 4 is $\frac{1}{12}.$

(ii) Let B be the event of getting numbers on top of dice with product 12.

B = {(2, 6), (3, 4), (4, 3), (6, 2)}.

∴ P(B) = $\frac{4}{36} = \frac{1}{9}.$

Hence, the probability that the product of the numbers on the top of two dice is 12 is $\frac{1}{9}.$

(iii) Let C be the event of getting numbers on top of dice with product 7.

C = {}.

∴ P(C) = $\frac{0}{36} = 0.$

Hence, the probability that the product of the numbers on the top of two dice is 7 is 0.