Probability

When a coin is tossed once, the possible outcomes are H and T.

Total number of possible outcomes = 2

(i) Favourable outcomes for getting a head = H

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Getting a head)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{2}\end{aligned}$$

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

(ii) Favourable outcomes for not getting a head = T

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Not getting a head)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{2}\end{aligned}$$

Hence, the probability of not getting a head is $\mathbf{\frac{1}{2}}$.

Total number of trials = 80

Number of times head is obtained = 38

Number of times tail is obtained = 80 − 38 = 42

(i) Number of favourable outcomes (Getting a tail) = 42

$$\begin{aligned}\text{P(Getting a tail)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{42}{80} \\[1em] = \frac{21}{40}\end{aligned}$$

Hence, the probability of getting a tail is $\mathbf{\frac{21}{40}}$.

(ii) Number of favourable outcomes (Getting a head) = 38

$$\begin{aligned}\text{P(Getting a head)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{38}{80} \\[1em] = \frac{19}{40}\end{aligned}$$

Hence, the probability of getting a head is $\mathbf{\frac{19}{40}}$.

Total number of outcomes = 20

(i) The outcome 3 appears 4 times.

Number of favourable outcomes (Getting 3) = 4

$$\begin{aligned}\text{P(Getting 3)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{4}{20} \\[1em] = \frac{1}{5}\end{aligned}$$

Hence, the probability of getting 3 is $\mathbf{\frac{1}{5}}$.

(ii) The numbers less than 3 are 1 and 2.

Number of times 1 or 2 appears = 2 + 3 = 5

Number of favourable outcomes (Getting a number less than 3) = 5

$$\begin{aligned}\text{P(Getting a number less than 3)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{5}{20} \\[1em] = \frac{1}{4}\end{aligned}$$

Hence, the probability of getting a number less than 3 is $\mathbf{\frac{1}{4}}$.

(iii) The numbers greater than 3 are 4, 5 and 6.

Number of times 4, 5 or 6 appears = 4 + 3 + 4 = 11

Number of favourable outcomes (Getting a number greater than 3) = 11

$$\begin{aligned}\text{P(Getting a number greater than 3)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{11}{20}\end{aligned}$$

Hence, the probability of getting a number greater than 3 is $\mathbf{\frac{11}{20}}$.

Total number of boys = 50

Number of boys who like tea = 21

Number of boys who dislike tea = 29

(i) Number of favourable outcomes (Choosing a boy who likes tea) = 21

$$\begin{aligned}\text{P(Choosing a boy who likes tea)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{21}{50}\end{aligned}$$

Hence, the probability that the chosen boy likes tea is $\mathbf{\frac{21}{50}}$.

(ii) Number of favourable outcomes (Choosing a boy who dislikes tea) = 29

$$\begin{aligned}\text{P(Choosing a boy who dislikes tea)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{29}{50}\end{aligned}$$

Hence, the probability that the chosen boy dislikes tea is $\mathbf{\frac{29}{50}}$.

Total number of balls played = 80

Number of times a boundary is hit = 12

Number of times a boundary is not hit = 80 − 12 = 68

(i) Number of favourable outcomes (Hitting a boundary) = 12

$$\begin{aligned}\text{P(Hitting a boundary)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{12}{80} \\[1em] = \frac{3}{20}\end{aligned}$$

Hence, the probability that he hits a boundary is $\mathbf{\frac{3}{20}}$.

(ii) Number of favourable outcomes (Not hitting a boundary) = 68

$$\begin{aligned}\text{P(Not hitting a boundary)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{68}{80} \\[1em] = \frac{17}{20}\end{aligned}$$

Hence, the probability that he does not hit a boundary is $\mathbf{\frac{17}{20}}$.

The marbles are numbered 1, 2, 3, 4, 5, 6, 7 and 8.

Total number of possible outcomes = 8

(i) Favourable outcomes for drawing the marble numbered 3 = 3

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Drawing the marble numbered 3)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{8}\end{aligned}$$

Hence, the probability of drawing a marble with number 3 is $\mathbf{\frac{1}{8}}$.

(ii) Favourable outcomes for drawing the marble numbered 7 = 7

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Drawing the marble numbered 7)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{8}\end{aligned}$$

Hence, the probability of drawing a marble with number 7 is $\mathbf{\frac{1}{8}}$.

Total number of trials = 100

(i) Number of times two heads are obtained = 21

$$\begin{aligned}\text{P(Getting two heads)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{21}{100}\end{aligned}$$

Hence, the probability of getting two heads is $\mathbf{\frac{21}{100}}$.

(ii) Number of times exactly one head is obtained = 55

$$\begin{aligned}\text{P(Getting exactly one head)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{55}{100} \\[1em] = \frac{11}{20}\end{aligned}$$

Hence, the probability of getting exactly one head is $\mathbf{\frac{11}{20}}$.

(iii) Number of times no head is obtained = 24

$$\begin{aligned}\text{P(Getting no head)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{24}{100} \\[1em] = \frac{6}{25}\end{aligned}$$

Hence, the probability of getting no head is $\mathbf{\frac{6}{25}}$.

Total number of balls = 4 + 6 = 10

(i) Number of favourable outcomes (Drawing a black ball) = 6

$$\begin{aligned}\text{P(Drawing a black ball)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{6}{10} \\[1em] = \frac{3}{5}\end{aligned}$$

Hence, the probability that the ball drawn is black is $\mathbf{\frac{3}{5}}$.

(ii) Number of favourable outcomes (Drawing a white ball) = 4

$$\begin{aligned}\text{P(Drawing a white ball)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{10} \\[1em] = \frac{2}{5}\end{aligned}$$

Hence, the probability that the ball drawn is white is $\mathbf{\frac{2}{5}}$.

(iii) A ball which is not black must be white.

Number of favourable outcomes (Drawing a ball which is not black) = 4

$$\begin{aligned}\text{P(Drawing a ball which is not black)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{10} \\[1em] = \frac{2}{5}\end{aligned}$$

Hence, the probability that the ball drawn is not black is $\mathbf{\frac{2}{5}}$.

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

(i) Favourable outcomes for getting 4 = 4

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Getting 4)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{6}\end{aligned}$$

Hence, the probability of getting 4 is $\mathbf{\frac{1}{6}}$.

(ii) Favourable outcomes for getting 6 = 6

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Getting 6)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{6}\end{aligned}$$

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

(iii) The numbers greater than 4 are 5 and 6.

Number of favourable outcomes (Getting a number greater than 4) = 2

$$\begin{aligned}\text{P(Getting a number greater than 4)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{6} \\[1em] = \frac{1}{3}\end{aligned}$$

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

Total number of cards = 100

Since the cards are numbered from 1 to 100, there is exactly one card bearing each number.

(i) Favourable outcomes for drawing card number 50 = 50

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Drawing card number 50)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{100}\end{aligned}$$

Hence, the probability of drawing card number 50 is $\mathbf{\frac{1}{100}}$.

(ii) Favourable outcomes for drawing card number 80 = 80

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Drawing card number 80)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{100}\end{aligned}$$

Hence, the probability of drawing card number 80 is $\mathbf{\frac{1}{100}}$.

(iii) Favourable outcomes for drawing card number 40 = 40

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Drawing card number 40)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{100}\end{aligned}$$

Hence, the probability of drawing card number 40 is $\mathbf{\frac{1}{100}}$.

(i) The complementary event S′ is that it will not snow tomorrow.

Hence, S′ is the event that it will not snow tomorrow.

(ii) We know that,

P(S) + P(S′) = 1

Given, P(S) = 0.03

P(S′) = 1 - P(S)

= 1 - 0.03

= 0.97

Hence, P(S′) = 0.97.

The given probabilities are:

A → 20%, B → 22%, C → 7%, D → 15% and E → 36%.

(i) Since E has the greatest probability, E is most likely to win the race.

Hence, E is most likely to win the race.

(ii) Since C has the least probability, C is least likely to win the race.

Hence, C is least likely to win the race.

(iii) Sum of the given probabilities = 20% + 22% + 7% + 15% + 36%

= 100%

= 1

Hence, the sum of the given probabilities is 100%, i.e., 1.

(iv) Since only one student can win the race, the events that A wins and D wins are mutually exclusive.

P(A or D wins) = 20% + 15% = 35%

Hence, the probability that either A or D will win the race is 35%, i.e., 0.35.

(v) Let S be the event that B will win the race.

(a) P(S) = 22% = 0.22

Hence, P(S) = 0.22.

(b) The complementary event S′ is that B will not win the race.

Hence, S′ is the event that B will not win the race.

(c) We know that,

P(S) + P(S′) = 1

P(S′) = 1 - P(S)

= 1 - 0.22

= 0.78

Hence, P(S′) = 0.78, i.e., 78%.

Total number of tickets = 3 + 4 + 5 = 12

(a) Number of favourable outcomes (Getting a green ticket) = 3

$$\begin{aligned}\text{P(Getting a green ticket)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{3}{12} \\[1em] = \frac{1}{4}\end{aligned}$$

Hence, the probability of getting a green ticket is $\mathbf{\frac{1}{4}}$.

(b) Number of green or yellow tickets = 3 + 4 = 7

Number of favourable outcomes (Getting a green or yellow ticket) = 7

$$\begin{aligned}\text{P(Getting a green or yellow ticket)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{7}{12}\end{aligned}$$

Hence, the probability of getting a green or yellow ticket is $\mathbf{\frac{7}{12}}$.

(c) There is no orange ticket in the basket.

Number of favourable outcomes (Getting an orange ticket) = 0

$$\begin{aligned}\text{P(Getting an orange ticket)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{0}{12} = 0\end{aligned}$$

Hence, the probability of getting an orange ticket is 0.

Total number of cards = 10

(a) Favourable outcomes for choosing 7 = 7

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Choosing 7)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{10}\end{aligned}$$

Hence, the probability of choosing 7 is $\mathbf{\frac{1}{10}}$.

(b) Favourable outcomes for choosing 9 or 10 = 9, 10

No. of favourable outcomes = 2

$$\begin{aligned}\text{P(Choosing 9 or 10)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{10} \\[1em] = \frac{1}{5}\end{aligned}$$

Hence, the probability of choosing 9 or 10 is $\mathbf{\frac{1}{5}}$.

(c) The numbers greater than 4 are 5, 6, 7, 8, 9 and 10.

Number of favourable outcomes (Choosing a number greater than 4) = 6

$$\begin{aligned}\text{P(Choosing a number greater than 4)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{6}{10} \\[1em] = \frac{3}{5}\end{aligned}$$

Hence, the probability of choosing a number greater than 4 is $\mathbf{\frac{3}{5}}$.

(d) The numbers less than 6 are 1, 2, 3, 4 and 5.

Number of favourable outcomes (Choosing a number less than 6) = 5

$$\begin{aligned}\text{P(Choosing a number less than 6)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{5}{10} \\[1em] = \frac{1}{2}\end{aligned}$$

Hence, the probability of choosing a number less than 6 is $\mathbf{\frac{1}{2}}$.

Total number of eggs = 8 + 4 = 12

(a) Number of favourable outcomes (Selecting a brown egg) = 8

$$\begin{aligned}\text{P(Selecting a brown egg)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{8}{12} \\[1em] = \frac{2}{3}\end{aligned}$$

Hence, the probability of selecting a brown egg is $\mathbf{\frac{2}{3}}$.

(b) Number of favourable outcomes (Selecting a white egg) = 4

$$\begin{aligned}\text{P(Selecting a white egg)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{12} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, the probability of selecting a white egg is $\mathbf{\frac{1}{3}}$.

Total number of tickets = 3 + 4 + 8 = 15

(a) Number of favourable outcomes (Getting a yellow ticket) = 3

$$\begin{aligned}\text{P(Getting a yellow ticket)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{3}{15} \\[1em] = \frac{1}{5}\end{aligned}$$

Hence, the probability of getting a yellow ticket is $\mathbf{\frac{1}{5}}$.

(b) Number of favourable outcomes (Getting a green ticket) = 4

$$\begin{aligned}\text{P(Getting a green ticket)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{15}\end{aligned}$$

Hence, the probability of getting a green ticket is $\mathbf{\frac{4}{15}}$.

(c) Number of favourable outcomes (Getting a blue ticket) = 8

$$\begin{aligned}\text{P(Getting a blue ticket)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{8}{15}\end{aligned}$$

Hence, the probability of getting a blue ticket is $\mathbf{\frac{8}{15}}$.

(d) There is no red ticket in the box.

Number of favourable outcomes (Getting a red ticket) = 0

$$\begin{aligned}\text{P(Getting a red ticket)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{0}{15} = 0\end{aligned}$$

Hence, the probability of getting a red ticket is 0.

(e) Number of tickets which are not yellow = 4 + 8 = 12

Number of favourable outcomes (Getting a ticket which is not yellow) = 12

$$\begin{aligned}\text{P(Getting a ticket which is not yellow)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{12}{15} \\[1em] = \frac{4}{5}\end{aligned}$$

Hence, the probability of getting a ticket which is not yellow is $\mathbf{\frac{4}{5}}$.

Number of males = 8 + 12 + 6 = 26

Number of females = 6 + 10 + 4 = 20

Total number of persons = 26 + 20 = 46

(a) Number of favourable outcomes (Picking a male) = 26

$$\begin{aligned}\text{P(Picking a male)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{26}{46} \\[1em] = \frac{13}{23}\end{aligned}$$

Hence, the probability that the person picked is a male is $\mathbf{\frac{13}{23}}$.

(b) Number of favourable outcomes (Picking a female) = 20

$$\begin{aligned}\text{P(Picking a female)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{20}{46} \\[1em] = \frac{10}{23}\end{aligned}$$

Hence, the probability that the person picked is a female is $\mathbf{\frac{10}{23}}$.

(c) Number of females aged 21–50 years = 10

$$\begin{aligned}\text{P(Picking a female aged 21–50 years)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{10}{46} \\[1em] = \frac{5}{23}\end{aligned}$$

Hence, the probability that the person picked is a female aged 21–50 years is $\mathbf{\frac{5}{23}}$.

(d) Number of males aged up to 50 years = 8 + 12 = 20

$$\begin{aligned}\text{P(Picking a male aged up to 50 years)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{20}{46} \\[1em] = \frac{10}{23}\end{aligned}$$

Hence, the probability that the person picked is a male aged up to 50 years is $\mathbf{\frac{10}{23}}$.

Total number of trials = 400

Number of times head is obtained = 270

Number of times tail is obtained = 400 − 270 = 130

(a) Number of favourable outcomes (Getting a head) = 270

$$\begin{aligned}\text{P(Getting a head)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{270}{400} \\[1em] = \frac{27}{40}\end{aligned}$$

Hence, the probability of getting a head is $\mathbf{\frac{27}{40}}$.

(b) Number of favourable outcomes (Getting a tail) = 130

$$\begin{aligned}\text{P(Getting a tail)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{130}{400} \\[1em] = \frac{13}{40}\end{aligned}$$

Hence, the probability of getting a tail is $\mathbf{\frac{13}{40}}$.

Total number of trials = 8

Number of times 5 is obtained = 2

$$\begin{aligned}\text{P(Getting 5)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{2}{8} \\[1em] = \frac{1}{4}\end{aligned}$$

Hence, the probability of getting 5 is $\mathbf{\frac{1}{4}}$.

In a single roll of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

(a) Favourable outcomes for getting 3 = 3

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Getting 3)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{6}\end{aligned}$$

Hence, the probability of getting 3 is $\mathbf{\frac{1}{6}}$.

(b) Favourable outcomes for getting 5 = 5

No. of favourable outcomes = 1

$$\begin{aligned}\text{P(Getting 5)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{6}\end{aligned}$$

Hence, the probability of getting 5 is $\mathbf{\frac{1}{6}}$.

(c) Favourable outcomes for getting 2 or 4 = 2, 4

No. of favourable outcomes = 2

$$\begin{aligned}\text{P(Getting 2 or 4)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{6} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, the probability of getting 2 or 4 is $\mathbf{\frac{1}{3}}$.

(d) The numbers more than 4 are 5 and 6.

Number of favourable outcomes (Getting a number more than 4) = 2

$$\begin{aligned}\text{P(Getting a number more than 4)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{6} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, the probability of getting a number more than 4 is $\mathbf{\frac{1}{3}}$.

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

The numbers greater than 4 are 5 and 6.

Number of favourable outcomes = 2

$$\begin{aligned}\text{P(Getting a number greater than 4)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{6} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, option 3 is the correct option.

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

There is no number greater than 6 on a die.

Number of favourable outcomes = 0

$$\begin{aligned}\text{P(Getting a number greater than 6)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{0}{6} = 0\end{aligned}$$

Hence, option 4 is the correct option.

Total number of balls = 5 + 8 + 7 = 20

Number of favourable outcomes (Getting a blue ball) = 8

$$\begin{aligned}\text{P(Getting a blue ball)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{8}{20} \\[1em] = \frac{2}{5}\end{aligned}$$

Hence, option 1 is the correct option.

An impossible event has no favourable outcome.

No. of favourable outcomes = 0

$$\begin{aligned}\text{P(Impossible event)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{0}{\text{Total number of possible outcomes}} = 0\end{aligned}$$

Therefore, the probability of an impossible event is 0.

Hence, option 2 is the correct option.

The word MUSSOORIE has 9 letters: M, U, S, S, O, O, R, I and E.

Total number of possible outcomes = 9

The vowels are U, O, O, I and E.

Number of favourable outcomes (Selecting a vowel) = 5

$$\begin{aligned}\text{P(Selecting a vowel)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{5}{9}\end{aligned}$$

Hence, option 4 is the correct option.

In one throw of a die, only one number can appear.

Therefore, getting both 3 and 5 in the same throw is impossible.

Total number of possible outcomes = 6

No. of favourable outcomes = 0

$$\begin{aligned}\text{P(Getting 3 and 5)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{0}{6} = 0\end{aligned}$$

Hence, option 2 is the correct option.

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

Favourable outcomes for getting 3 or 5 = 3, 5

No. of favourable outcomes = 2

$$\begin{aligned}\text{P(Getting 3 or 5)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{6} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, option 1 is the correct option.

The square numbers from 1 to 50 are 1, 4, 9, 16, 25, 36 and 49.

Total number of possible outcomes = 50

Number of favourable outcomes (Drawing a square number) = 7

$$\begin{aligned}\text{P(Drawing a square number)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{7}{50}\end{aligned}$$

Hence, option 3 is the correct option.

The months whose names start with J are January, June and July.

Total number of possible outcomes = 12

Number of favourable outcomes = 3

$$\begin{aligned}\text{P(Getting the name of a month starting with J)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{3}{12} \\[1em] = \frac{1}{4}\end{aligned}$$

Hence, option 1 is the correct option.

Total number of outcomes = 100

The numbers less than 4 are 1, 2 and 3.

Number of times 1, 2 or 3 appears = 20 + 15 + 15 = 50

Number of favourable outcomes = 50

$$\begin{aligned}\text{P(Getting a number less than 4)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{50}{100} \\[1em] = \frac{1}{2}\end{aligned}$$

Hence, option 2 is the correct option.

For Statement 1:

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

There is no prime number greater than 6 among these outcomes.

Total number of possible outcomes = 6

No. of favourable outcomes = 0

$$\begin{aligned}\text{P(Getting a prime number greater than 6)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{0}{6} = 0\end{aligned}$$

Therefore, getting a prime number greater than 6 is an impossible event and its probability is 0.

So, Statement 1 is true.

For Statement 2:

An event which has no chance of occurring is called an impossible event.

So, Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

For Statement 1:

In a random experiment, the possible outcomes are known, but it is not known in advance which one of them will occur.

Therefore, Statement 1 is false.

For Statement 2:

The probability of an event lies between 0 and 1, both inclusive. Hence, it can also be expressed as a percentage from 0% to 100%.

Therefore, Statement 2 is true.

∴ Statement 1 is false, and Statement 2 is true.

Hence, option 4 is the correct option.

The first 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20.

There is no odd number among them.

Therefore,

$$\begin{aligned}\text{P(Getting an odd number)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{0}{10} = 0\end{aligned}$$

So, Assertion (A) is false.

The probability of an event A is given by

$\text{P(A)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

So, Reason (R) is true.

∴ A is false and R is true.

Hence, option 2 is the correct option.

The probabilities of solving the problem are:

P → 32%, Q → 36% and R → 32%.

Since 36% is the greatest probability, Q is most likely to solve the problem.

So, Assertion (A) is true.

Events having equal probabilities are called equally likely events. They are not necessarily the most likely events.

So, Reason (R) is false.

∴ A is true and R is false.

Hence, option 1 is the correct option.

Assuming that the chances of a newly born baby being a girl or a boy are equal,

$$\begin{aligned}\text{P(A baby is a girl)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \\[1em] = \frac{1}{2} = 0.5\end{aligned}$$

So, Assertion (A) is true.

If the chances of happening and non-happening of an event are the same, the event is said to have even chances of happening.

So, Reason (R) is true.

∴ Both A and R are true.

Hence, option 3 is the correct option.