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.

Q1.

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4 . What is the probability that the student knows the answer given that he answered it correctly

Opt.

Let E1 = Student knows the answer
E2 = Student guesses the answer
And A = answer is correct

P(E1)=34P(A | E1)=1P(E2)=14P(A | E2)=14Thus, by Bayes’s theorem,P(E1 | A)=P(E1)P(A | E1)P(E1

Ans.

Let E1 = Student knows the answer
E2 = Student guesses the answer
And A = answer is correct

P(E1)=34P(A | E1)=1P(E2)=14P(A | E2)=14Thus, by Bayes’s theorem,P(E1 | A)=P(E1)P(A | E1)P(E1)P(A | E1)+P(E2)P(A | E2)=34—134—1+14—14=1213.

Q2.

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

Opt.

Let S1 = six occurs

S2 = six does not occur.

and E = the man reports that six occurs in the throwing of the die

P(S1)=16P(E | S1)=34P(S2)=56P(E | S2)=14Using Bayes’s theorem,P(S1 | E)=P(S1)P(E | S1)

Ans.

Let S1 = six occurs

S2 = six does not occur.

and E = the man reports that six occurs in the throwing of the die

P(S1)=16P(E | S1)=34P(S2)=56P(E | S2)=14Using Bayes’s theorem,P(S1 | E)=P(S1)P(E | S1)P(S1)P(E | S1)+P(S2)P(E | S2)=16—3416—34+56—14=38.

Q3.

Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.

Opt.

Let E1 = choosing the bag I, E2 = choosing the bag II  and A = drawing a red ball

P(E1)=12P(A | E1)=P (drawing a red ball from bag I) =37P(E2)=12P(A | E2)=P (drawing a red ball from bag II) =511Using Bayes’s theorem,

Ans. Let E1 = choosing the bag I, E2 = choosing the bag II and A = drawing a red ball.

P(E1)=12P(A | E1)=P (drawing a red ball from bag I) =37P(E2)=12P(A | E2)=P (drawing a red ball from bag II) =511Using Bayes’s theorem,P(E2 | A)=P(E2)P(A | E2)P(E1)P(A | E1) + P(E2)P(A | E2)                =12—51122—37+12—511                =3568.

 

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