 # Important Questions for Class 12 Maths chapter 13 Probability

Probability Chapter 13 of Class 12 Maths is one of the most important topics in CBSE Board Exams. The important questions for class 12 Maths Chapter 13 – Probability will help you to practice well for exams. Solving these questions will help you in being confident about the topic. Also, get all the important questions for class 12 Maths chapters in BYJU’S website.

After solving these important questions on Probability, you can practise the given NCERT Questions too. These questions of Chapter 13 class 12 contains all types of questions – short and long answers. This topic is also important for preparing for entrance exams. In this article, you are going to have a look at the important questions for class 12 maths chapter 13 probability with the solutions.

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## Class 12 Chapter 13 Probability Important Questions with Solutions

Practice the given solved problems for the board examinations. Solving these problems will help you to increase the efficiency of writing solutions in the final exam.

Q.1:  A die is thrown twice and the sum of the numbers rising is noted to be 6. Calculate the is 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 atleast 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, subsbtitute the probability values obtained= (2/36)/ (5/36)

Hence, Required probability is 2/5.

Q.2:  Given that the events A and B are such that P(A) = 1/2, P (A ∪ B) = 3/5, and P(B) = p. Find p if they are

(i) mutually exclusive

(ii) independent

Solution:

Given, P(A) = 1/2 ,

P (A ∪ B) = 3/5

and P(B) = p.

(1) For Mutually Exclusive

Given that, the sets A and B are mutually exclusive.

Thus, they do not have any common elements

Therefore, P(A ∩ B) = 0

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Substitute the formulas in the above-given formula, we get

3/5 = (1/2) + p – 0

Simplify the expression, we get

(3/5) – (1/2) = p

(6 − 5)/10 = p

1/10 = p

Therefore, p = 1/10

Hence, the value of p is 1/10, if they are mutually exclusive.

(ii) For Independent events:

If the two events A & B are independent,

we can write it as P(A ∩ B) = P(A) P(B)

Substitute the values,

= (1/2) × p

= p/2

Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Now, substitute the values in the formula,

(3/5) = (1/2)+ p – (p/2)

(3/2)– (1/2)= p – (p/2)

(6 − 5)/10 = p/2

1/10 = p/2

p= 2/10

P = 1/5

Thus, the value of p is 1/5, if they are independent.

Q.3: The probability of solving the specific problem independently by the persons’ A and B are 1/2 and 1/3 respectively. In case, if both the persons 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.

Q. 4: 5 cards are drawn successively from a well-shuffled pack of 52 cards with replacement. Determine the probability that (i) all the five cards should be spades? (ii) only 3 cards should be spades? (iii) none of the cards is a spade?

Solution:

Let us assume that X be the number of spade cards

Using the Bernoulli trial, X has a binomial distribution

P(X = x) = nCx qn-x px

Thus, the number of cards drawn, n = 5

Probability of getting spade card, p = 13/52 = 1/4

Thus the value of the q can be found using

q = 1 – p = 1 – (1/4)= 3/4

Now substitute the p and q values in the formula,

Hence, P(X = x) = 5Cx (3/4)5-x(1/4)x

(1) Probability of Getting all the spade cards:

P(all the five cards should be spade) = 5𝐶5 (1/4)5(3/4)0

= (1/4)5

= 1/1024

(2) Probability of Getting only three spade cards:

P(only three cards should be spade) = 5𝐶3 (1/4)3(3/4)2

= (5!/3! 2!) × (9/1024)

= 45/ 512

(3) Probability of Getting no spades:

P(none of the cards is a spade) = 5𝐶0(1/4)0(3/4)5

= (3/4)5

= 243/ 1024

Q. 5: An 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.

### Practice Problems for Class 12 Maths Chapter 13 Probability

1) Two numbers are selected at random from the integers 1 through 9. If the sum is even, find the probability that both the numbers are odd. (Answer: 5/8).

2) A die is rolled. If the outcome is an odd number, what is the probability that it is prime? (Answer: 2/3).

3) A coin is tossed twice. If the outcome is at most one tail, what is the probability that both head and tail have appeared? (Answer: 2/3).

4) An unbiased die is tossed twice. Find the probability of getting a 4, 5, 6on the first toss and a 1, 2, 3, 4 on the second toss. (Answer: 1/3).

5) The probability that person A hits a target is 1/3 and the probability that person B hits it is 2/5. What is the probability that the target will be hit if both person A and person B shoot at it? (Answer: 3/5).