Probability

ICSE Class 7 Mathematics · Solutions for Class 7 RS Aggarwal ICSE Foundation Mathematics · 7 solved questions

Exercise

1Exercise 25

When a coin was tossed 200 times, we obtained

Head : 118 times and Tail : 82 times.

When the coin is tossed at random, what is the probability of getting

(i) a head?

(ii) a tail?

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Total number of trials = 200.

Number of heads obtained = 118

Number of tails obtained = 82

(i) P (getting a head) = $\frac{\text{no. of heads in these trials}}{\text{total no. of trials}}$

$$\begin{aligned}= \frac{118}{200} \\[1em] = \frac{59}{100}\end{aligned}$$

∴ Probability of getting a head = $\frac{59}{100}$

(ii) P (getting a tail) = $\frac{\text{no. of tails in these trials}}{\text{total no. of trials}}$

$$\begin{aligned}= \frac{82}{200} \\[1em] = \frac{41}{100}\end{aligned}$$

∴ Probability of getting a tail = $\frac{41}{100}$

2Exercise 25

A die is rolled 150 times and the outcomes are noted which are tabulated as shown below :

Outcome	1	2	3	4	5	6
Frequency	29	17	24	19	33	28

When a die is rolled at random, find the probability of getting a :

(i) 1

(ii) 4

(iii) 5

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Total number of trials = 150.

(i) Number of times 1 is obtained = 29.

P (getting 1) = $\frac{\text{no. of times 1 is obtained}}{\text{total no. of trials}}$

= $\frac{29}{150}$

∴ Probability of getting 1 = $\frac{29}{150}$

(ii) Number of times 4 is obtained = 19.

P (getting 4) = $\frac{\text{no. of times 4 is obtained}}{\text{total no. of trials}}$

= $\frac{19}{150}$.

∴ Probability of getting 4 = $\frac{19}{150}$

(iii) Number of times 5 is obtained = 33.

P (getting 5) = $\frac{\text{no. of times 5 is obtained}}{\text{total no. of trials}}$

$$\begin{aligned}= \frac{33}{150} \\[1em] = \frac{11}{50}\end{aligned}$$

∴ Probability of getting 5 = $\frac{11}{50}$

3Exercise 25

Five cards numbered 1 to 5 are placed in a bag. A card is drawn from the bag 30 times. Each time, the number on the drawn card is noted and the card drawn is replaced. The outcomes are tabulated as shown below :

Outcome	1	2	3	4	5
Frequency	4	10	7	5	4

If a card is now drawn at random, find the probability that the card drawn bears the number :

(i) 2

(ii) 3

(iii) 5

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Total number of trials = 30.

(i) Number of times 2 is obtained = 10.

P (getting 2) = $\frac{\text{no. of times 2 is obtained}}{\text{total no. of trials}}$

$$\begin{aligned}= \frac{10}{30} \\[1em] = \frac{1}{3}\end{aligned}$$

∴ Probability of getting 2 = $\frac{1}{3}$

(ii) Number of times 3 is obtained = 7.

P (getting 3) = $\frac{\text{no. of times 3 is obtained}}{\text{total no. of trials}}$

= $\frac{7}{30}$.

∴ Probability of getting 3 = $\frac{7}{30}$

(iii) Number of times 5 is obtained = 4.

P (getting 5) = $\frac{\text{no. of times 5 is obtained}}{\text{total no. of trials}}$

$$\begin{aligned}= \frac{4}{30} \\[1em] = \frac{2}{15}\end{aligned}$$

∴ Probability of getting 5 = $\frac{2}{15}$

4Exercise 25

Two coins are tossed simultaneously 100 times. We obtained :

Two Heads : 23 times; One Head : 56 times; Zero Head : 21 times.

When two coins are tossed at random, what is the probability of getting:

(i) Two Heads?

(ii) One Head?

(iii) Zero Head?

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Total number of trials = 100.

(i) Number of times two heads obtained = 23.

P (getting two heads) = $\frac{\text{no. of times two heads obtained}}{\text{total no. of trials}}$

= $\frac{23}{100}$.

∴ Probability of getting two heads = $\frac{23}{100}$

(ii) Number of times one head obtained = 56.

P (getting one head) = $\frac{\text{no. of times one head obtained}}{\text{total no. of trials}}$

$$\begin{aligned}= \frac{56}{100} \\[1em] = \frac{14}{25}\end{aligned}$$

∴ Probability of getting one head = $\frac{14}{25}$

(iii) Number of times zero head obtained = 21.

P (getting zero head) = $\frac{\text{no. of times zero head obtained}}{\text{total no. of trials}}$

= $\frac{21}{100}$.

∴ Probability of getting zero head = $\frac{21}{100}$

5Exercise 25

In a survery of 200 employees of a company, it was found that 89 liked the lunch packs provided by the company while the rest did not. Out of these employees, if one employee is chosen at random, what is the probability that the chosen employee:

(i) likes the lunch pack?

(ii) dislikes the lunch pack?

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Total number of employees surveyed = 200.

Number of employees who liked the lunch packs = 89.

Number of employees who disliked the lunch packs = 200 − 89 = 111.

(i) P (chosen employee likes the lunch packs) = $\frac{\text{No. of employees who liked the lunch packs}}{\text{Total no. of employees surveyed}}$

= $\frac{89}{200}$

∴ Probability that the chosen employee likes the lunch packs = $\frac{89}{200}$

(ii) P (chosen employee dislikes the lunch packs) = $\frac{\text{No. of employees who disliked the lunch packs}}{\text{Total no. of employees surveyed}}$

= $\frac{111}{200}$

∴ Probability that the chosen employee dislikes the lunch packs = $\frac{111}{200}$

1Exercise 25 - Assertions and Reasons

Assertion : A coin is tossed 16 times and the outcomes are recorded as below :

H T T H T H H H T T H T H T T H

The probability of occurrence of a head is 50%.

Reason : When a coin is tossed, there are two possible outcomes—Head and Tail.

(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c)Assertion (A) is true but Reason (R) is false.
(d)Assertion (A) is false but Reason (R) is true.

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Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation

Total number of trials = 16.

Number of times head occurs = 8.

∴ P (getting a head) = $\frac{\text{no. of times head occurs}}{\text{total no. of trials}}$

$$\begin{aligned}= \frac{8}{16} \\[1em] = \frac{1}{2} \\[1em]\end{aligned}$$

= 50%

So, the Assertion is true.

When a coin is tossed, the two possible outcomes are Head and Tail.

So, the Reason is also true.

However, the Reason only states the possible outcomes when a coin is tossed and does not explain how the empirical probability of getting a head is calculated from the given data. Hence, R is not the correct explanation of A.

Hence, option 2 is the correct option.

2Exercise 25 - Assertions and Reasons

Assertion : When a spinner with three colours Red (R), Green (G) and Black (B) as shown is rotated, red and green colours have equal probability to show up with arrow.

When a spinner with three colours Red (R), Green (G) and Black (B) as shown is rotated, red and green colours have equal probability to show up with arrow. Mensuration, Foundation Mathematics R.S. Aggarwal ICSE Class 7.

Reason : The probability of occurrence of an event E is given by

$\text{P(E)}=\frac{\text{total no. of trials}}{\text{no. of trials in which E occurs}}$

(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c)Assertion (A) is true but Reason (R) is false.
(d)Assertion (A) is false but Reason (R) is true.

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

Explanation

From the spinner, Red (R) and Green (G) cover equal regions of the circle. So, when the spinner is rotated, red and green colours have equal probability to show up with the arrow.

Hence, the Assertion is true.

The correct formula for the probability of occurrence of an event E is :

$\text{P(E)}=\frac{\text{no. of trials in which E occurs}}{\text{total no. of trials}}$

The formula given in the Reason is inverted, so the Reason is false.

Hence, option 3 is the correct option.

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