Master Class 10 Maths Chapter 14 Probability Ex 14.2 CBSE

Introduction:

In this exercise Class 10 Maths Chapter 14 Probability Ex 14.2 (CBSE) 15.2 (SEBA), we will explore the basic principles of probability, which helps us calculate the likelihood of an event occurring. Probability is a branch of mathematics that deals with the study of random events and the chances of their occurrence.

This exercise involves problems based on the calculation of probability using different methods, including the use of equally likely outcomes. We will work with situations like drawing balls from a bag, tossing coins, rolling dice, and other experiments to calculate the probability of different events. Understanding the concepts in this chapter helps students develop a strong foundation in probability, which is an essential topic in mathematics.

Class 10 Maths Chapter 14 Probability Ex 14.2 (CBSE) 15.2 (SEBA)

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: Given:

Shyam and Ekta are visiting a shop from Tuesday to Saturday, which gives 5 possible days for each of them to visit.
Each customer is equally likely to visit the shop on any day, and their visits are independent of each other.

Total number of possible outcomes:
Since each customer can visit the shop on any of the 5 days, the total number of possible outcomes for both customers visiting the shop is:
5 × 5 = 25

(i) Probability that both will visit the shop on the same day:

For both customers to visit the shop on the same day, there are 5 possible favorable outcomes:

Shyam and Ekta both visit on Tuesday, both on Wednesday, both on Thursday, both on Friday, or both on Saturday.

So, the probability is:
P(same day) = 5/25 = 1/5

(ii) Probability that both will visit on consecutive days:

For both to visit on consecutive days, there are 4 possible favorable outcomes for each day:

If Shyam visits on Tuesday, Ekta can visit on Wednesday.
If Shyam visits on Wednesday, Ekta can visit on Thursday.
If Shyam visits on Thursday, Ekta can visit on Friday.
If Shyam visits on Friday, Ekta can visit on Saturday.

So, the number of favorable outcomes is 4.

The probability is:
P(consecutive days) = 4/25

(iii) Probability that both will visit on different days:

For both customers to visit on different days, we first calculate the number of favorable outcomes for different days. Since there are 25 total outcomes, and 5 outcomes where they visit the same day, and 4 outcomes where they visit consecutive days, the remaining outcomes are for them visiting on different days.

The number of favorable outcomes is:
Different days = 25 – 5 – 4 = 16

So, the probability is:
P(different days) = 16/25

Final Answers:
(i) The probability that both will visit the shop on the same day is 1/5.
(ii) The probability that both will visit on consecutive days is 4/25.
(iii) The probability that both will visit on different days is 16/25.

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 twice, and the total score in the two throws is noted. Complete the following table, which gives a few values of the total score on the two throws:

Number in first throw	1	2	2	3	3	6
Number in second throw
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
6	7	8	9	10	11	12

The table shows the sum of the numbers for different combinations of throws. Now, let’s solve the probability questions:

(i) What is the probability that the total score is even?

To find the probability of getting an even total score, let’s first count the total number of outcomes for the two throws. The die has 6 faces, so for each throw, there are 6 outcomes. Therefore, the total number of outcomes is:
6 × 6 = 36

Next, count the number of outcomes where the total score is even. From the table, the even total scores are:

2 (from 1 + 1)
4 (from 1 + 3, 2 + 2)
6 (from 1 + 5, 2 + 4, 3 + 3)
8 (from 2 + 6, 3 + 5)
10 (from 3 + 7)
12 (from 6 + 6)

There are 18 outcomes where the total score is even. The probability is:
P(even total score) = 18/36 = 1/2

(ii) What is the probability that the total score is at least 6?

To find the probability of getting a total score of at least 6, let’s count the number of outcomes where the total score is 6 or greater. From the table, the outcomes where the total score is at least 6 are:

6 (from 1 + 5, 2 + 4, 3 + 3)
7 (from 1 + 6, 2 + 5, 3 + 4)
8 (from 2 + 6, 3 + 5)
9 (from 3 + 6)
10 (from 3 + 7)
11 (from 6 + 5)
12 (from 6 + 6)

There are 21 outcomes where the total score is at least 6. The probability is:
P(at least 6) = 21/36 = 7/12

Final Answers:
(i) The probability that the total score is even is 1/2.
(ii) The probability that the total score is at least 6 is 7/12.

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is doubles that of a red ball, determine the number of blue balls in the bag.
Solution:
Given:

The bag contains 5 red balls.
Let the number of blue balls be xxx.
The probability of drawing a blue ball is double the probability of drawing a red ball.

Step 1: Probability of drawing a red ball
The total number of balls in the bag is 5+x5 + x5+x (since there are 5 red balls and xxx blue balls).
The probability of drawing a red ball is:
P(red) = 5 / (5 + x)

Step 2: Probability of drawing a blue ball
The probability of drawing a blue ball is:
P(blue) = x / (5 + x)

Step 3: Given condition
We are told that the probability of drawing a blue ball is twice the probability of drawing a red ball:
P(blue) = 2 × P(red)
Substitute the expressions for P(blue) and P(red):
x / (5 + x) = 2 × 5 / (5 + x)

Step 4: Solve the equation
Since the denominators are the same, we can cancel them out:
x = 2 × 5
x = 10

Final Answer:
The number of blue balls in the bag is 10.

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:
Given:

The box contains 12 balls, of which xxx are black.
The probability of drawing a black ball is given.
If 6 more black balls are added to the box, the probability of drawing a black ball becomes double what it was before.

Step 1: Initial probability of drawing a black ball
The total number of balls is 12, and the number of black balls is xxx.
The probability of drawing a black ball initially is:
P(black) = x / 12

Step 2: Probability after adding 6 black balls
After adding 6 more black balls, the total number of balls becomes 12+6=1812 + 6 = 1812+6=18, and the number of black balls becomes x+6x + 6x+6.
The new probability of drawing a black ball is:
P(new black) = (x + 6) / 18

Step 3: Given condition
The new probability is double the initial probability. So, we have:
P(new black) = 2 × P(black)
Substitute the expressions for P(new black) and P(black):
(x + 6) / 18 = 2 × x / 12

Step 4: Solve the equation
Simplify the equation:
(x + 6) / 18 = x / 6
Cross multiply to eliminate the denominators:
6(x + 6) = 18x
Expand both sides:
6x + 36 = 18x
Simplify:
36 = 18x – 6x
36 = 12x
Solve for x:
x = 36 / 12 = 3

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In conclusion, understanding probability allows us to predict the likelihood of different events occurring in random experiments. By calculating the probability of an event, we gain insight into how likely or unlikely certain outcomes are. This exercise has helped us reinforce the concepts of probability through practical examples such as rolling dice and drawing balls. By mastering these concepts, we can confidently apply probability to solve real-life problems. Understanding probability enables us to make informed predictions and better comprehend the randomness of various situations.