NCERT Solutions - Probability JEE Notes | EduRev

Mathematics (Maths) Class 12

JEE : NCERT Solutions - Probability JEE Notes | EduRev

The document NCERT Solutions - Probability JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Exercise 13.1

Question 1:

Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E).

Answer 1: 

It is given that P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2 

NCERT Solutions - Probability JEE Notes | EduRev

Question 2:

Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

Answer 2: 

It is given that P(B) = 0.5 and P(A ∩ B) = 0.32

NCERT Solutions - Probability JEE Notes | EduRev

Question 3:

If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find

(i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)

Answer 3: 

It is given that P(A) = 0.8, P(B) = 0.5, and P(B|A) = 0.4

(i) P (B|A) = 0.4

NCERT Solutions - Probability JEE Notes | EduRev
NCERT Solutions - Probability JEE Notes | EduRev

Question 4:

Evaluate P (A ∪ B), if 2P (A) = P (B) = 5/13 and P(A|B) = 2/5

Answer 4:

NCERT Solutions - Probability JEE Notes | EduRev
NCERT Solutions - Probability JEE Notes | EduRev

Question 5:

If P(A) = 6/11, P(B) =5/11 and P(A ∪ B) =7/11 , find

(i) P(A ∩ B) (ii) P(A|B) (iii) P(B|A)

Answer 5:  

It is given that NCERT Solutions - Probability JEE Notes | EduRev

NCERT Solutions - Probability JEE Notes | EduRev

NCERT Solutions - Probability JEE Notes | EduRev
NCERT Solutions - Probability JEE Notes | EduRev
NCERT Solutions - Probability JEE Notes | EduRev

Question 6:

A coin is tossed three times, where

(i) E: head on third toss, F: heads on first two tosses

(ii) E: at least two heads, F: at most two heads

(iii) E: at most two tails, F: at least one tail

Answer 6: 

If a coin is tossed three times, then the sample space S is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space has 8 elements.

(i) E = {HHH, HTH, THH, TTH}

F = {HHH, HHT}

NCERT Solutions - Probability JEE Notes | EduRevE ∩ F = {HHH}

NCERT Solutions - Probability JEE Notes | EduRev

(ii) E = {HHH, HHT, HTH, THH}

F = {HHT, HTH, HTT, THH, THT, TTH, TTT}

NCERT Solutions - Probability JEE Notes | EduRevE ∩ F = {HHT, HTH, THH}

NCERT Solutions - Probability JEE Notes | EduRev

(iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH}

F = {HHT, HTT, HTH, THH, THT, TTH, TTT}

NCERT Solutions - Probability JEE Notes | EduRev

Question 7:

Two coins are tossed once, where

(i) E: tail appears on one coin, F: one coin shows head

(ii) E: not tail appears, F: no head appears

Answer 7: 

If two coins are tossed once, then the sample space S is

S = {HH, HT, TH, TT}

(i) E = {HT, TH}

F = {HT, TH}

NCERT Solutions - Probability JEE Notes | EduRev

(ii) E = {HH}

F = {TT}

∴ E ∩ F = Φ

P (F) = 1 and P (E ∩ F) = 0

∴ P(E|F) =  NCERT Solutions - Probability JEE Notes | EduRev

Question 8:

A die is thrown three times,

E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses

Answer 8: 

If a die is thrown three times, then the number of elements in the sample space will be 6 × 6 × 6 = 216

NCERT Solutions - Probability JEE Notes | EduRev
NCERT Solutions - Probability JEE Notes | EduRev

Question 9:

Mother, father and son line up at random for a family picture

E: son on one end, F: father in middle

Answer 9: 

If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be

S = {MFS, MSF, FMS, FSM, SMF, SFM}

⇒ E = {MFS, FMS, SMF, SFM}

F = {MFS, SFM}

∴ E ∩ F = {MFS, SFM}

NCERT Solutions - Probability JEE Notes | EduRev  

Question 10:

A black and a red dice are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Answer 10: 

Let the first observation be from the black die and second from the red die.

When two dice (one black and another red) are rolled, the sample space S has 6 × 6 = 36 number of elements.
 Let A: Obtaining a sum greater than 9

= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}

B: Black die results in a 5.

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

∴ A ∩ B = {(5, 5), (5, 6)}

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B).

NCERT Solutions - Probability JEE Notes | EduRev

(b) E: Sum of the observations is 8.

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

F: Red die resulted in a number less than 4.

NCERT Solutions - Probability JEE Notes | EduRev

The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E|F). 

NCERT Solutions - Probability JEE Notes | EduRev

Question 11:

A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}

Find (i) P (E|F) and P (F|E) (ii) P (E|G) and P (G|E) (ii) P ((E ∪ F)|G) and P ((E ∩ G)|G)

Answer 11: 

When a fair die is rolled, the sample space S will be

S = {1, 2, 3, 4, 5, 6}

It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5}

NCERT Solutions - Probability JEE Notes | EduRev
NCERT Solutions - Probability JEE Notes | EduRev

(ii) E ∩ G = {3, 5}

NCERT Solutions - Probability JEE Notes | EduRev

(iii) E ∪ F = {1, 2, 3, 5}

(E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5}

E ∩ F = {3}

(E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3}

NCERT Solutions - Probability JEE Notes | EduRev
NCERT Solutions - Probability JEE Notes | EduRev

Question 12:

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Answer 12: 
 Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be

S = {(bb), (bg), (gb), (g, g)}

Let A be the event that both children are girls.

NCERT Solutions - Probability JEE Notes | EduRev

(i) Let B be the event that the youngest child is a girl.

NCERT Solutions - Probability JEE Notes | EduRev

The conditional probability that both are girls, given that the youngest child is a girl, is given by P (A|B).

NCERT Solutions - Probability JEE Notes | EduRev

Therefore, the required probability is (1/2).

(ii) Let C be the event that at least one child is a girl.

NCERT Solutions - Probability JEE Notes | EduRev

The conditional probability that both are girls, given that at least one child is a girl, is given by P(A|C).

NCERT Solutions - Probability JEE Notes | EduRev

Question 13:

An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Answer 13: 

The given data can be tabulated as

 

True/False

Multiple choice

Total

Easy

300

500

800

Difficult

200

400

600

Total

500

900

1400

 

 

 

 

 

 

 

Let us denote E = easy questions, M = multiple choice questions, D = difficult questions, and T = True/False questions

Total number of questions = 1400

Total number of multiple choice questions = 900

Therefore, probability of selecting an easy multiple choice question is

P (E ∩ M) = 500/1400= 5/14

Probability of selecting a multiple choice question, P (M), is 900/1400= 9/14

P (E|M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.

NCERT Solutions - Probability JEE Notes | EduRev

Therefore, the required probability is 5/9

Question 14:

Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Answer 14: 
 When dice is thrown, number of observations in the sample space = 6 × 6 = 36

Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different.
 ∴ A = {(1, 3), (2, 2), (3, 1)}

NCERT Solutions - Probability JEE Notes | EduRev

Let P (A|B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.

Question 15:

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Answer 15: 

The outcomes of the given experiment can be represented by the following tree diagram.

The sample space of the experiment is,

NCERT Solutions - Probability JEE Notes | EduRev

Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.

NCERT Solutions - Probability JEE Notes | EduRev
NCERT Solutions - Probability JEE Notes | EduRev

Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by P(A|B).

Therefore, 

NCERT Solutions - Probability JEE Notes | EduRev

Question 16:

If  NCERT Solutions - Probability JEE Notes | EduRev 
 (A) 0 (B) (1/2)

(C) not defined (D) 1

Answer 16:
NCERT Solutions - Probability JEE Notes | EduRev

Therefore, P (A|B) is not defined.

Thus, the correct answer is C.

Question 17:

If A and B are events such that P (A|B) = P(B|A), then

(A) A ⊂ B but A ≠ B (B) A = B

(C) A ∩ B = Φ (D) P(A) = P(B)

Answer 17: 

It is given that, P(A|B) = P(B|A) 

NCERT Solutions - Probability JEE Notes | EduRev

Thus, the correct answer is D.

Exercise 13.2

Question 1:

NCERT Solutions - Probability JEE Notes | EduRev   find P (A ∩ B) if A and B are independent events.

Answer 1:
 It is given that NCERT Solutions - Probability JEE Notes | EduRev

A and B are independent events. Therefore,

NCERT Solutions - Probability JEE Notes | EduRev

Question 2:

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Answer 2: 

There are 26 black cards in a deck of 52 cards.

Let P (A) be the probability of getting a black card in the first draw.

P(A) = 26/52 = 1/2

Let P (B) be the probability of getting a black card on the second draw.

Since the card is not replaced,

P(B) =25/51

Thus, probability of getting both the cards black =  NCERT Solutions - Probability JEE Notes | EduRev

Question 3:

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Answer 3: 

Let A, B, and C be the respective events that the first, second, and third drawn orange is good.

Therefore, probability that first drawn orange is good, P(A) = 12/15

The oranges are not replaced.

Therefore, probability of getting second orange good, P(B) = 11/14

Similarly, probability of getting third orange good, P(C) = 10/3

The box is approved for sale, if all the three oranges are good.

Thus, probability of getting all the oranges good  NCERT Solutions - Probability JEE Notes | EduRev

Therefore, the probability that the box is approved for sale is 44/91.

Question 4:

A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

Answer 4: 

If a fair coin and an unbiased die are tossed, then the sample space S is given by,

NCERT Solutions - Probability JEE Notes | EduRev
NCERT Solutions - Probability JEE Notes | EduRev

Therefore, A and B are independent events.

Question 5:

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?

Answer 5: 

When a die is thrown, the sample space (S) is

S = {1, 2, 3, 4, 5, 6}

Let A: the number is even = {2, 4, 6}

NCERT Solutions - Probability JEE Notes | EduRev

B: the number is red = {1, 2, 3}

NCERT Solutions - Probability JEE Notes | EduRev

Therefore, A and B are not independent.

Question 6:

Let E and F be events with NCERT Solutions - Probability JEE Notes | EduRev Are E and F independent?

Answer 6: It is given that NCERT Solutions - Probability JEE Notes | EduRev

NCERT Solutions - Probability JEE Notes | EduRev

Therefore, E and F are not independent.

Question 7:

Given that the events A and B are such thatNCERT Solutions - Probability JEE Notes | EduRev  and P (B) = p. Find p if they are (i) mutually exclusive (ii) independent.

It is given that NCERT Solutions - Probability JEE Notes | EduRev

(i) When A and B are mutually exclusive, A ∩ B = Φ

∴ P (A ∩ B) = 0

NCERT Solutions - Probability JEE Notes | EduRev

(ii) When A and B are independent, NCERT Solutions - Probability JEE Notes | EduRev

NCERT Solutions - Probability JEE Notes | EduRev

Question 8:

Let A and B be independent events with P (A) = 0.3 and P (B) = 0.4. Find

(i) P (A ∩ B) (ii) P (A ∪ B)

(iii) P (A|B) (iv) P (B|A)

Answer 8: 

It is given that P (A) = 0.3 and P (B) = 0.4

(i) If A and B are independent events, then 

NCERT Solutions - Probability JEE Notes | EduRev

NCERT Solutions - Probability JEE Notes | EduRev 

Question 9:

If A and B are two events such that NCERT Solutions - Probability JEE Notes | EduRev find P (not A and not B).

Answer 9: 

It is given that, NCERT Solutions - Probability JEE Notes | EduRev

P(not on A and not on B) = NCERT Solutions - Probability JEE Notes | EduRev

P(not on A and not on B) = NCERT Solutions - Probability JEE Notes | EduRev

NCERT Solutions - Probability JEE Notes | EduRev

Question 10:

Events A and B are such that NCERT Solutions - Probability JEE Notes | EduRev

State whether A and B are independent?

Answer 10: 

It is given that NCERT Solutions - Probability JEE Notes | EduRev

NCERT Solutions - Probability JEE Notes | EduRev

Therefore, A and B are not independent events. 

Question 11:

Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find

(i) P (A and B) (ii) P (A and not B)

(iii) P (A or B) (iv) P (neither A nor B)

Answer 11: 

It is given that P (A) = 0.3 and P (B) = 0.6

Also, A and B are independent events.

Also, A and B are independent events.

NCERT Solutions - Probability JEE Notes | EduRev
NCERT Solutions - Probability JEE Notes | EduRev

Question 12:

A die is tossed thrice. Find the probability of getting an odd number at least once.

Answer 12: 

Probability of getting an odd number in a single throw of a die 3/6= 1/2 

Similarly, probability of getting an even number = 3/6 = 1/2

Probability of getting an even number three times =  NCERT Solutions - Probability JEE Notes | EduRev

Therefore, probability of getting an odd number at least once

= 1 − Probability of getting an odd number in none of the throws 

= 1 − Probability of getting an even number thrice

=1-1/8= 7/8

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