PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.2

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 15 Probability Ex 15.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Solution:
When Shyam and Ekta visit a particular shop in the same week. Possible outcomes are:
S = {(T, T) (T, W) (T, Th) (T, F) (T, S) (W, T) (W, W) (W, Th) (W, F) (W, S) (Th, T) (Th, W) (Th, Th) (Th, F) (Th, S) (F, T) (F, W) (F, Th) (F, F) (F, S) (S, T) (S, W) (S, Th) (S, F) (S, S)}

Here T stands For Tuesday
W stands For Wednesday
Th stands For Thursday
F stands For Friday
S stands For Saturday
n(S) = 25
(i) Let A is event that Shyam and Ekta visit the shop on the same day
A = {(T, T) (W, W) (Th, Th) (F, F) (S, S)}
n(A) = 5
Probability that both will visit the shop on same day = \(\frac{5}{25}\)
∴ P(A) = \(\frac{1}{5}\).

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.2

(ii) Let B is event that both will visit consecutive days particular shop
B = {(T, W) (W, T) (W, Th) (Th, W) (Th, F) (F, Th) (F, S) (F, S)}
n(B) = 8
∴ Probability that both will visit particular shop on consecutive days = \(\frac{8}{25}\).

(iii) Probability that both will visit the shop on different days = 1 – Probability that both will visit the shop
on same day.
= 1 – \(\frac{1}{5}\) [∵ P \((\bar{A})\) = 1 – P(A)]
= \(\frac{5-1}{5}\)
P \((\bar{A})\) = \(\frac{4}{5}\).

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.2

Question 2.
A die, is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.2 1

What is the probability that the total score is
(i) even?
(ii) 6?
(iii) at least 6?
Solution:
The complete table is

PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.2 2

Number of all possible out comes = 6 × 6 = 36
(i) Let A is event of getting total as even
A = {2, 4, 4, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 8, 8, 8, 8, 12}
n (A) = 18
∴ Probability of getting an even number = \(\frac{18}{36}=\frac{1}{2}\)
P (Even Number) = \(\frac{1}{2}\).

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.2

(ii) Let B is event of getting sum as 6 B = {6, 6, 6, 6)
n(B) = 4
Probability of getting an even number = \(\frac{4}{36}\)
∴ P(B) = \(\frac{1}{9}\).

(iii) Let C is event of getting sum at least 6
C = (6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 12}
n(C) = 15
∴ Probability of getting at least 6 = \(\frac{15}{36}=\frac{5}{12}\)
∴ P(C) = \(\frac{5}{12}\).

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.2

Question 3.
A bag contains 5 red balls and some blue balls. lithe probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution:
Number of red balls = 5
Let number of blue balls = x
∴ Total number of balls = 5 + x
According to question,
Probability of drawing blue ball = 2 Probability of Red ball
\(\frac{x}{5+x}=2\left[\frac{5}{5+x}\right]\)
x = 10
∴ Number of blue balls = 10.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.2

Question 4.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution:
Total number of balls in bag = 12
Number of black balls x
∴ Probability of getting black ball = \(\frac{x}{12}\)
If 6 more balls put in the box then total number of balls in the box = 12 + 6 = 18
Number of black balls = x + 6
Probability of getting black ball = \(\frac{x+6}{18}\)
According to Question,
Probability of drawing black ball = 2
Probability of drawing blackball in first case
\(\frac{x+6}{18}=\frac{2 x}{12}\)

\(\frac{x+6}{3}=\frac{2 x}{2}\)

\(\frac{x+6}{3}\) = x
x + 6 = 3x
6 = 3x – x
6 = 2x
x = 3
∴ Number of black balls = 3.

PSEB Solutions PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.2

Question 5.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{2}{3}\). Find the number of blue marbles in the jar.
Solution:
Total number of marbles in jar =24
Let number of green marbles = x
∴ Number of blue marbles = 24 – x
P (Green marbles) = \(\frac{x}{24}\)
When a marble is drawn
Probability of drawing green marble = \(\frac{2}{3}\) (Given)
\(\frac{x}{24}=\frac{2}{3}\)
x = \(\frac{24 \times 2}{3}\)
x = 16

∴ Number of green marbles = 16
∴ Number of blue marbles = 24 – x = 24 – 16 = 8.

Leave a Comment