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BSEB Class 10 Maths Chapter 15 Probability Ex 15.2 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 15 Probability Ex 15.2 Book Answers

BSEB Class 10 Maths Chapter 15 Probability Ex 15.2 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 15 Probability Ex 15.2 Book Answers
BSEB Class 10 Maths Chapter 15 Probability Ex 15.2 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 15 Probability Ex 15.2 Book Answers


BSEB Class 10th Maths Chapter 15 Probability Ex 15.2 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 10th Maths Chapter 15 Probability Ex 15.2 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 10th Maths Chapter 15 Probability Ex 15.2 books are published by the Bihar Board Publishers. These Bihar Board Class 10th Maths Chapter 15 Probability Ex 15.2 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 10th Maths Chapter 15 Probability Ex 15.2 book solutions pdf online from this page.

Bihar Board Class 10th Maths Chapter 15 Probability Ex 15.2 Textbooks Solutions PDF

Bihar Board STD 10th Maths Chapter 15 Probability Ex 15.2 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 10th Maths Chapter 15 Probability Ex 15.2 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 10th Maths Chapter 15 Probability Ex 15.2 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 10th Maths Chapter 15 Probability Ex 15.2 Textbooks. These Bihar Board Class 10th Maths Chapter 15 Probability Ex 15.2 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.

Bihar Board Class 10th Maths Chapter 15 Probability Ex 15.2 Books Solutions

Board BSEB
Materials Textbook Solutions/Guide
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Class 10th
Subject Maths Chapter 15 Probability Ex 15.2
Chapters All
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BSEB Class 10th Maths Chapter 15 Probability Ex 15.2 Textbooks Solutions with Answer PDF Download

Find below the list of all BSEB Class 10th Maths Chapter 15 Probability Ex 15.2 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

BSEB Bihar Board Class 10th Maths Solutions 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:
Possible outcomes associated to the random experiment of visiting a particular shop in the same week (Tuesday to Saturday) by two customers Shyam and Ekta are –
(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).
where T = Tuesday, W = Wednesday, Th = Thursday, F = Friday and S = Saturday.
∴ Total number of possible outcomes = 5 x 5 = 25

(i) Let A be the event of visiting a particular shop on the same day by two customers. Then, outcomes favourable to A are (T, T), (W, W), (TH, TH), (F, F) and (S, S)
∴ Favourable number of outcomes for same day = 5
Hence, required probability = 525 = 15.

(ii) Let A be the event of visiting a particular shop by two customers on consecutive days. Then, outcomes favourable to A are (T, W), (W, T), (W, TH), (TH, W), (TH, F), (F, TH), (S, F) and (F, S).
∴ Favourable number of outcomes for consecutive days = 8
Hence, the required probability = 525 = 15.

(iii) Let A be the event of visiting a particular shop by two customers on different days. Then, outcomes favourable to A are excluding (T, T), (W, W), (TH, TH), (F, F) and (S, S). So, these are 25 – 5 = 20 in number;
∴ Favourable number of outcomes for different days = 20
Hence, required probability = 2025 = 45

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 :

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

Clearly total number of possible outcomes = 6 x 6 = 36
(i) Let A be the event of getting total score even. Then, outcomes favourable to A are 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8 and 12
∴ Favourable number of outcomes for even numbers = 18
Hence, P(A) = 1836 = 12

(ii) Let A be the event of getting total score 6. Then, outcomes favourable to A are 6, 6, 6 and 6.
∴ Favourable number of outcomes for 6 = 4
Hence, P(A) = 436 = 19

(iii) Let A be the event of the total score is at least 6. Then, outcomes favourable to A are 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12.
∴ Favourable number of outcomes for at least 6 = 15
Hence, P(A) = 1536 = 512

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution:
Let there be x blue balls in the bag.
∴ Total number of balls in the bag = 5 + x
Now, p1 = Probability of drawing a blue ball
= 𝑥5+𝑥
p2 = Probability of drawing a red ball = 55+𝑥
But it is given that p1 = 2p2.
So, 𝑥5+𝑥 = 2 x 55+𝑥 or x = 10
Hence, there are 10 blue balls in the bag.

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:
There are 12 balls in the box. Out of these 12 balls, one can be chosen in 12 ways.
∴ Total number of possible outcomes = 12
There are x black balls out of which one can be chosen ir x ways.
∴ Favourable number of outcomes for black ball = x
Hence, p1 = P (getting a black ball) = 𝑥12.
If 6 more black balls are put in the box, then
Total number of balls in the box = 12 + 6 = 18
Number of black balls in the box = x + 6
∴ p2 = P (getting a black ball) = 𝑥+618
It is given that p2 = 2p1
So, 𝑥+618 = 2 x 𝑥12
or 𝑥+618 = 𝑥6
or x + 6 = 3x or 2x = 6 or x – 3

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 23. Find the number of blue marbles in the jar.
Solution:
There are 24 marbles in the jar, some are green and others are blue.
Total number of possible outcomes = 24 Let there be x green marbles.
∴ Favourable number of outcomes for green marbles = x
∴ P(G) = 𝑥24 But, P(G) = 23 [Given]
So, 𝑥24 = 23 or x = 23 x 24 = 16
or Number of green marbles = 16
∴ Number of blue marbles = 24 – 16 = 8.


BSEB Textbook Solutions PDF for Class 10th


Bihar Board Class 10th Maths Chapter 15 Probability Ex 15.2 Textbooks for Exam Preparations

Bihar Board Class 10th Maths Chapter 15 Probability Ex 15.2 Textbook Solutions can be of great help in your Bihar Board Class 10th Maths Chapter 15 Probability Ex 15.2 exam preparation. The BSEB STD 10th Maths Chapter 15 Probability Ex 15.2 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 15 Probability Ex 15.2 Books State Board syllabus with maximum efficiency.

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