Important Questions Class 12 Maths Chapter 13

Important Questions Class 12 Mathematics Chapter 13

Important Questions for CBSE Class 12 Mathematics Chapter 13 – Probability

Important Questions Class 12 Mathematics Chapter 13 Probability is a crucial topic in CBSE Board Exams. The Mathematics Class 12 Chapter 13 Important Questions – Probability will help you in exam preparation. On the Extramarks website, you can find all of the Class 12 Mathematics Chapter 13 Important Questions.

CBSE Class 12 Mathematics Chapter-13 Important Questions

Study Important Questions for Class 12 Mathematics Chapter 13 – Probability

Here are a few of the Important Questions from Class 12 Mathematics Chapter 13. For the full set, access the link given here.

Q1. A die is thrown twice and the sum of the numbers rising is noted to be 6. Calculate the conditional probability that the number 4 has arrived at least once?

Solution: If a dice is thrown twice, then the sample space obtained is:

S = {(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)

(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}

From the given data, it is needed to find the Probability that 4 has appeared at least once, given the sum of nos. is observed to be 6

Assume that, F: Addition of numbers is 6

and take E: 4 has appeared at least once

So, that, we need to find P(E|F)

Finding P (E):

The probability of getting 4 at least once is:

E = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}

Thus , P(E) = 11/ 36

Finding P (F):

The probability to get the addition of numbers is 6 is:

F = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)}

Thus, P(F) = 5/ 36

Also, E ∩ F = {(2,4), (4,2)}

P(E ∩ F) = 2/36

Thus, P(E|F) = (P(E ∩ F) ) / (P (F) )

Now, substitute the probability values obtained= (2/36)/ (5/36)

Hence, Required probability is 2/5.

Q2. The probability of solving the specific problem independently by the persons’ A and B are 1/2 and 1/3 respectively. In case both people try to solve the problem independently, then calculate the probability that the problem is solved.

Solution:

Given that, the two events say A and B are independent if P(A ∩ B) = P(A). P(B)

From the given data, we can observe that P(A) = 1/2 & P(B) = 1/3

The probability that the problem is solved = Probability that person A solves the problem or the person B solves the Problem

This can be written as:

= P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

If A and B are independent, then P(A ∩ B) = P(A). P(B)

Now, substitute the values,

= (1/2) × (1/3)

P(A ∩ B) = 1/6

Now, the probability of problem solved is written as

P(Problem is solved) = P(A) + P(B) – P(A ∩ B)

= (1/2) + (1/3) – (1/6)

= (3/6) + (2/6) – (1/6)

= 4/6

= 2/3

Hence, the probability of the problem solved is 2/3.

Q3. A fair die is thrown double times. Assume that the event A is “odd number on the first throw” and B the event “odd number on the second throw”. Compare the independence of the events A and B.

Solution:

Let us consider two independent events A and B, then P(A ∩ B) = P(A). P(B)

when an unbiased die is thrown twice

S = {(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)

(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}

Let us describe two events as

A: odd number on the first throw

B: odd number on the second throw

To find P(A)

A = {(1, 1), (1, 2), (1, 3), …, (1, 6)

(3, 1), (3, 2), (3, 3), …, (3, 6)

(5, 1), (5, 2), (5, 3), …, (5, 6)}

Thus, P (A) = 18/36 = 1/2

To find P(B)

B = {(1, 1), (2, 1), (3, 1), …, (6, 1)

(1, 3), (2, 3), (3, 3), …, (6, 3)

(1, 5), (2, 5), (3, 5), …, (6, 5)}

Thus, P (B) = 18/36 = 1/2

A ∩ B = odd number on the first & second throw = { (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}

So, P(A ∩ B) = 9/36 = 1/ 4

Now, P(A). P(B) = (1/2) × (1/2) = 1/4

As P(A ∩ B) = P(A). P(B),

Hence, the two events A and B are independent events.