NCERT Solutions for Class 10 Maths Chapter 15 - Probability

• Choose the correct answer from the given four options.

1. If an event cannot occur, then its probability is

(a) 1

$$\textbf{(b)\space}\frac{\textbf{3}}{\textbf{4}}$$

$$\textbf{(c)\space}\frac{\textbf{1}}{\textbf{2}}$$

(d) 0

Sol. (d) 0

2. Which of the following cannot be the probability of an event?

$$\textbf{(a)}\space\frac{\textbf{1}}{\textbf{3}}$$

(b) 0.1

(c) 3

$$\textbf{(d)}\space\frac{\textbf{17}}{\textbf{16}}\\\textbf{Sol.}\space\text{(d)}\space\frac{17}{16}$$

15. A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A, B and C is

$$\textbf{(a)\space}\frac{\textbf{4}}{\textbf{23}}$$

$$\textbf{(b)\space}\frac{\textbf{6}}{\textbf{23}}$$

$$\textbf{(c)\space}\frac{\textbf{8}}{\textbf{23}}$$

$$\textbf{(d)\space}\frac{\textbf{17}}{\textbf{23}}$$

$$\textbf{Sol.}\space\text{(b)}\space\frac{6}{23}$$

5. Two dice are numbered 1, 2, 3, 4, 5, ,6 and 1, 1, 2, 2, 3, 3 respectively. They are thrown and then sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 respectively.

Sol. Number of total outcomes = 6² = 36

(i) Let E₁ = Event of getting sum 2 = {(1, 1), (1, 1)}

n(E₁) = 2

$$\therefore\space\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}\\=\frac{2}{36}=\frac{1}{18}$$

(ii) E₂ = Event of getting sum 3 = {(1, 2), (1, 2), (2, 1), (2, 1)}

n(E₂) = 4

$$\therefore\space\text{P(E}_2)=\frac{n(\text{E}_2)}{\text{n(S)}}\\=\frac{4}{36}=\frac{1}{9}$$

(iii) Let E₃ = Event of getting sum 4 = {(2, 2), (2, 2), (3, 1), (3, 1), (1, 3), (1, 3)}

∴ n(E₃) = 6

$$\therefore\space\text{P(E}_3)=\frac{\text{n(E}_3)}{\text{n(S)}}\\=\frac{6}{36}=\frac{1}{6}$$

(iv) Let E₄ = Event of getting sum 5 = {(2, 3), (2, 3), (4, 1), (4, 1), (3, 2), (3, 2)}

∴ n(E₄) = 6

$$\therefore\space\text{P(E}_4)=\frac{\text{n(E}_4)}{\text{n(S)}}\\=\frac{6}{36}=\frac{1}{6}$$

(v) Let E₅ = Event of getting sum 6 = {(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)}

n(E₅) = 6

$$\therefore\space\text{P(E}_5)=\frac{\text{n(E}_5)}{\text{n(S)}}\\=\frac{6}{36}=\frac{1}{6}$$

(vi) Let E6 = Event of getting sum 7 = {(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)}

∴ n(E₆) = 6

$$\therefore\space\text{P(E}_6)=\frac{\text{n(E}_6)}{\text{n(S)}}\\=\frac{6}{36}=\frac{1}{6}$$

(vii) Let E₇ = Event of getting sum 8 = {(5, 3), (5, 3), (5, 2), (6, 2)}

∴ n(E₁) = 4

$$\therefore\space\text{P(E}_7)=\frac{\text{n(E}_7)}{\text{n(S)}}\\=\frac{4}{36}=\frac{1}{9}$$

(viii) Let E₈ = Event of getting sum 9 = {(6, 3), (6, 3)}

∴ n(E₈) = 2

$$\therefore\space\text{P(E}_8)=\frac{\text{n(E}_8)}{\text{n(S)}}\\=\frac{2}{36}=\frac{1}{18}$$

9. All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value.

(i) 7

(ii) greater than 7

(iii) less than 7

Sol. In out of 52 play cards, 4 jacks, 4 queens and 4 kings are removed, then the remaining cards are left, n(s) = 52 – 3 × 4 = 40.

(i) Let E₁ = Event of getting card value is 7.

E = Card value 7 has either a spade is diamond, a club or a heart

∴ n(E₁) = 4

$$\therefore\space\text{P(E}_1)=\frac{\text{n(E}_1)}{\text{n(S)}}\\=\frac{4}{40}=\frac{1}{10}$$

(ii) Let E₂ = Event of getting card value is 7.

E = Event of getting card value is 8, 9 and 10

∴ n(E₂) = 3 × 4

(∵ Each card value has four cards)

= 12

$$\therefore\space\text{P(E}_2)=\frac{\text{n(E}_2)}{\text{n(S)}}\\=\frac{12}{40}=\frac{3}{10}$$