Class 12 Mathematics - NCERT Mathematics Part II
NCERT Solutions for Class 12 Maths Chapter 13: Probability
Chapter 13 covers advanced probability concepts building on Class 11 foundations. Students learn conditional probability P(A|B), the multiplication theorem, independent events, total probability theorem, and Bayes' theorem. Applications include real-world scenarios like medical testing, manufacturing defects, and exam predictions. This chapter carries 8 marks as an independent unit in the CBSE Class 12 board exam 2027.
Syllabus Update 2026-27
As per the CBSE 2026-27 rationalised syllabus, Chapter 13 has 3 exercises plus a Miscellaneous Exercise. Exercise 13.1 covers conditional probability, Exercise 13.2 covers multiplication theorem and independent events, and Exercise 13.3 covers Bayes' theorem. Random variables and Bernoulli trials (formerly Ex 13.4 and 13.5) have been removed from the syllabus.
Table of Contents
13.1Exercise 13.1 - Conditional Probability
Q1Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E intersection F) = 0.2, find P(E|F) and P(F|E).
P(E|F) = P(E intersection F)/P(F) = 0.2/0.3 = 2/3
P(F|E) = P(E intersection F)/P(E) = 0.2/0.6 = 1/3
Q2Compute P(A|B), if P(B) = 0.5 and P(A intersection B) = 0.32.
P(A|B) = P(A intersection B)/P(B) = 0.32/0.5 = 0.64 = 16/25
Q3If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find (i) P(A intersection B) (ii) P(A|B) (iii) P(A union B)
(i) P(B|A) = P(A intersection B)/P(A)
0.4 = P(A intersection B)/0.8
P(A intersection B) = 0.32
(ii) P(A|B) = P(A intersection B)/P(B) = 0.32/0.5 = 0.64
(iii) P(A union B) = P(A) + P(B) - P(A intersection B) = 0.8 + 0.5 - 0.32 = 0.98
Q4Evaluate P(A union B), if 2P(A) = P(B) = 5/13 and P(A|B) = 2/5.
2P(A) = 5/13 implies P(A) = 5/26
P(B) = 5/13
P(A|B) = P(A intersection B)/P(B)
2/5 = P(A intersection B)/(5/13)
P(A intersection B) = (2/5)(5/13) = 2/13
P(A union B) = P(A) + P(B) - P(A intersection B)
= 5/26 + 5/13 - 2/13
= 5/26 + 10/26 - 4/26
= 11/26
Q5If P(A) = 6/11, P(B) = 5/11 and P(A union B) = 7/11, find (i) P(A intersection B) (ii) P(A|B) (iii) P(B|A)
(i) P(A intersection B) = P(A) + P(B) - P(A union B) = 6/11 + 5/11 - 7/11 = 4/11
(ii) P(A|B) = P(A intersection B)/P(B) = (4/11)/(5/11) = 4/5
(iii) P(B|A) = P(A intersection B)/P(A) = (4/11)/(6/11) = 4/6 = 2/3
Q6A coin is tossed three times. Find P(E|F) 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
Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, |S|=8
(i) E = {HHH, HTH, THH, TTH} (head on 3rd), F = {HHH, HHT} (heads on 1st two)
E intersection F = {HHH}
P(E|F) = P(E intersection F)/P(F) = (1/8)/(2/8) = 1/2
(ii) E = {HHH, HHT, HTH, THH} (>=2 heads), F = {HHT, HTH, HTT, THH, THT, TTH, TTT} (<=2 heads)
E intersection F = {HHT, HTH, THH}
P(E|F) = (3/8)/(7/8) = 3/7
(iii) E = {HHH, HHT, HTH, THH, HTT, THT, TTH} (<=2 tails), F = {HHT, HTH, HTT, THH, THT, TTH, TTT} (>=1 tail)
E intersection F = {HHT, HTH, THH, HTT, THT, TTH}
P(E|F) = (6/8)/(7/8) = 6/7
Q7Two coins are tossed once. Find P(E|F) where E: no tail appears, F: no head appears.
S = {HH, HT, TH, TT}
E = {HH} (no tail), F = {TT} (no head)
E intersection F = empty set
P(E|F) = P(E intersection F)/P(F) = 0/(1/4) = 0
E and F are mutually exclusive, so P(E|F) = 0.
Q8A die is thrown three times. E: 4 appears on the third throw, F: 6 and 5 appears respectively on first two throws. Find P(E|F).
When a die is thrown 3 times, total outcomes = 6^3 = 216.
F = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}, |F| = 6
E intersection F = {(6,5,4)}, |E intersection F| = 1
P(E|F) = (1/216)/(6/216) = 1/6
This makes sense: given the first two throws are fixed, the third throw is independent with P(4) = 1/6.
Q9Mother, father and son line up at random for a family photograph. E: son on one end, F: father in middle. Find P(E|F).
All arrangements: {MFS, MSF, FMS, FSM, SMF, SFM}, total = 3! = 6
E (son on one end) = {SMF, SFM, MFS, FMS} → |E| = 4
F (father in middle) = {MFS, SFM} → |F| = 2
E intersection F = {MFS, SFM} → |E intersection F| = 2
P(E|F) = (2/6)/(2/6) = 1
If father is in the middle, son is always on an end (since there are only 3 people).
Q10A black and a red die are rolled. (a) Find P(A|B) where A: sum=8, B: red die shows 4. (b) Find P(A|B) where A: sum>=8, B: red die shows number<4.
(a) Total outcomes = 36.
B (red=4) = {(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)}, |B|=6
A (sum=8) and B: need black+4=8, black=4. So A intersection B = {(4,4)}, |A intersection B|=1
P(A|B) = (1/36)/(6/36) = 1/6
(b) B (red<4) = {all (x,y) where y=1,2,3}, |B|=18
A (sum>=8) and B: enumerate pairs where x+y>=8 and y<4:
y=1: x>=7 (none), y=2: x>=6→(6,2), y=3: x>=5→(5,3),(6,3)
A intersection B = {(6,2),(5,3),(6,3)}, |A intersection B|=3
P(A|B) = (3/36)/(18/36) = 3/18 = 1/6
Q11A 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) (iii) P((E union F)|G) and P((E intersection F)|G)
(i) E intersection F = {3}. P(E|F) = (1/6)/(2/6) = 1/2. P(F|E) = (1/6)/(3/6) = 1/3.
(ii) E intersection G = {3,5}. P(E|G) = (2/6)/(4/6) = 1/2. P(G|E) = (2/6)/(3/6) = 2/3.
(iii) E union F = {1,2,3,5}. (E union F) intersection G = {2,3,5}. P((E union F)|G) = (3/6)/(4/6) = 3/4.
E intersection F = {3}. (E intersection F) intersection G = {3}. P((E intersection F)|G) = (1/6)/(4/6) = 1/4.
Q12Assume 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?
S = {BB, BG, GB, GG} (first child listed first)
(i) A = both girls = {GG}. B = youngest is girl = {BG, GG}.
A intersection B = {GG}.
P(A|B) = (1/4)/(2/4) = 1/2
(ii) A = {GG}. C = at least one girl = {BG, GB, GG}.
A intersection C = {GG}.
P(A|C) = (1/4)/(3/4) = 1/3
Q13An 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?
Total questions = 300 + 200 + 500 + 400 = 1400
A = easy question, B = MCQ
P(B) = (500+400)/1400 = 900/1400
P(A intersection B) = 500/1400 (easy MCQs)
P(A|B) = P(A intersection B)/P(B) = (500/1400)/(900/1400) = 500/900 = 5/9
Q14Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4'.
Total outcomes = 36. F = two numbers different = 36 - 6 = 30 outcomes.
E = sum is 4 = {(1,3),(2,2),(3,1)}. E intersection F = {(1,3),(3,1)} (excluding (2,2) since digits same).
P(E|F) = (2/36)/(30/36) = 2/30 = 1/15
Q15Consider 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 P(E|F) where E: 'the coin shows a tail', F: 'at least one die shows a 3'.
If die shows 3 or 6 (multiples of 3): throw die again → outcomes (3,1),(3,2),...,(3,6),(6,1),...,(6,6) = 12 outcomes
If die shows 1,2,4,5: toss coin → outcomes (1,H),(1,T),(2,H),(2,T),(4,H),(4,T),(5,H),(5,T) = 8 outcomes
Total sample space = 20 outcomes (not equally likely from original, but each original die outcome has prob 1/6, then second has 1/6 or 1/2)
F = at least one die shows 3 = {(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)} = 6 outcomes (all from first branch)
E = coin shows tail = {(1,T),(2,T),(4,T),(5,T)} = 4 outcomes
E intersection F = empty set (F involves dice only, E involves coin)
P(E|F) = 0
Q16If P(A) = 1/2, P(B) = 0, then P(A|B) is (A) 0 (B) 1/2 (C) not defined (D) 1
P(A|B) = P(A intersection B)/P(B) = ?/0
Since P(B) = 0, P(A|B) is not defined (division by zero).
Answer: (C) not defined
Q17If A and B are events such that P(A|B) = P(B|A), then (A) A is a subset of B but A != B (B) A = B (C) A intersection B = empty (D) P(A) = P(B)
P(A|B) = P(B|A)
P(A intersection B)/P(B) = P(A intersection B)/P(A)
If P(A intersection B) != 0, then 1/P(B) = 1/P(A), so P(A) = P(B).
If P(A intersection B) = 0, then both sides are 0, which holds for any P(A), P(B).
But in general, the condition implies P(A) = P(B).
Answer: (D) P(A) = P(B)
13.2Exercise 13.2 - Multiplication Theorem and Independent Events
Q1If P(A) = 3/5 and P(B) = 1/5, find P(A intersection B) if A and B are independent events.
For independent events: P(A intersection B) = P(A) * P(B)
= (3/5)(1/5) = 3/25
Q2Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
P(1st black) = 26/52 = 1/2
P(2nd black | 1st black) = 25/51
P(both black) = (26/52)(25/51) = (1/2)(25/51) = 25/102
Q3A 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.
P(1st good) = 12/15 = 4/5
P(2nd good | 1st good) = 11/14
P(3rd good | first two good) = 10/13
P(all 3 good) = (12/15)(11/14)(10/13) = (4/5)(11/14)(10/13)
= 41110 / (51413) = 440/910 = 44/91
Q4A 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.
P(A) = 1/2 (head on coin)
P(B) = 1/6 (3 on die)
P(A intersection B) = P(head and 3) = (1/2)(1/6) = 1/12
P(A) * P(B) = (1/2)(1/6) = 1/12 = P(A intersection B) ✓
Since P(A intersection B) = P(A) * P(B), events A and B are independent.
Q5A die is thrown. If E is the event 'the number appearing is a multiple of 3' and F be the event 'the number appearing is even' then find whether E and F are independent.
E = {3, 6}, P(E) = 2/6 = 1/3
F = {2, 4, 6}, P(F) = 3/6 = 1/2
E intersection F = {6}, P(E intersection F) = 1/6
P(E) * P(F) = (1/3)(1/2) = 1/6 = P(E intersection F) ✓
E and F are independent.
Q6An unbiased die is thrown twice. Let event A be 'odd number on the first throw' and B be 'odd number on the second throw'. Check the independence of the events A and B.
P(A) = 3/6 = 1/2 (odd on 1st throw)
P(B) = 3/6 = 1/2 (odd on 2nd throw)
P(A intersection B) = (3/6)(3/6) = 9/36 = 1/4
P(A) * P(B) = (1/2)(1/2) = 1/4 = P(A intersection B) ✓
A and B are independent. (The two throws are independent experiments.)
Q7Three coins are tossed simultaneously. Consider E: 'three heads or three tails', F: 'at least two heads', G: 'at most two heads'. Of the pairs (E,F), (E,G) and (F,G), which are independent? which are dependent?
S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
E = {HHH, TTT}, P(E) = 2/8 = 1/4
F = {HHH,HHT,HTH,THH}, P(F) = 4/8 = 1/2
G = all except HHH = {HHT,HTH,HTT,THH,THT,TTH,TTT}, P(G) = 7/8
E intersection F = {HHH}, P = 1/8. P(E)*P(F) = (1/4)(1/2) = 1/8 ✓ E,F independent.
E intersection G = {TTT}, P = 1/8. P(E)*P(G) = (1/4)(7/8) = 7/32 ≠ 1/8. E,G dependent.
F intersection G = {HHT,HTH,THH}, P = 3/8. P(F)*P(G) = (1/2)(7/8) = 7/16 ≠ 3/8. F,G dependent.
Q8Two events A and B will be independent, if (A) A and B are mutually exclusive (B) P(A'B') = [1-P(A)][1-P(B)] (C) P(A) = P(B) (D) P(A)+P(B) = 1
For independent A,B: P(A intersection B) = P(A)*P(B)
(B): P(A'B') = P(A' intersection B') = P((A union B)') = 1 - P(A union B)
= 1 - [P(A)+P(B)-P(A intersection B)]
If independent: = 1 - P(A) - P(B) + P(A)P(B) = [1-P(A)][1-P(B)] ✓
Conversely, if P(A'B') = [1-P(A)][1-P(B)], we can show independence.
Answer: (B)
Q9If A and B are two events such that P(A) = 1/4, P(B) = 1/2 and P(A intersection B) = 1/8, find P(not A and not B).
P(A' intersection B') = P((A union B)') = 1 - P(A union B)
= 1 - [P(A) + P(B) - P(A intersection B)]
= 1 - [1/4 + 1/2 - 1/8]
= 1 - [2/8 + 4/8 - 1/8]
= 1 - 5/8
= 3/8
Q10Events A and B are such that P(A) = 1/2, P(B) = 7/12 and P(not A or not B) = 1/4. State whether A and B are independent.
P(A' union B') = P((A intersection B)') = 1 - P(A intersection B) = 1/4
So P(A intersection B) = 1 - 1/4 = 3/4
For independence: P(A)*P(B) = (1/2)(7/12) = 7/24
But P(A intersection B) = 3/4 = 18/24 ≠ 7/24
A and B are NOT independent.
Note: P(A intersection B) = 3/4 > P(A) = 1/2, which means P(B|A) > 1 - this is impossible. The given data appears inconsistent. In NCERT, the answer states they are not independent.
Q11Given 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)
Since A and B are independent:
(i) P(A and B) = P(A)*P(B) = 0.3 * 0.6 = 0.18
(ii) P(A and B') = P(A)*P(B') = 0.3 * 0.4 = 0.12
(iii) P(A or B) = P(A) + P(B) - P(A and B) = 0.3 + 0.6 - 0.18 = 0.72
(iv) P(A' and B') = P(A')*P(B') = 0.7 * 0.4 = 0.28
Q12A die is tossed thrice. Find the probability of getting an odd number at least once.
P(odd on single throw) = 3/6 = 1/2
P(even on single throw) = 1/2
P(odd at least once) = 1 - P(no odd in 3 throws)
= 1 - (1/2)^3
= 1 - 1/8
= 7/8
Q13Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red (ii) first is black and second is red (iii) one of them is black and other is red.
Total balls = 18. With replacement, draws are independent.
(i) P(both red) = (8/18)(8/18) = (4/9)(4/9) = 16/81
(ii) P(1st black, 2nd red) = (10/18)(8/18) = (5/9)(4/9) = 20/81
(iii) P(one black, one red) = P(BR) + P(RB)
= (10/18)(8/18) + (8/18)(10/18) = 2 * 20/81 = 40/81
Q14Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.
(i) P(solved) = 1 - P(neither solves) = 1 - P(A')P(B')
= 1 - (1/2)(2/3) = 1 - 2/6 = 1 - 1/3 = 2/3
(ii) P(exactly one) = P(A)P(B') + P(A')P(B)
= (1/2)(2/3) + (1/2)(1/3)
= 2/6 + 1/6 = 3/6 = 1/2
Q15One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent? (i) E: 'the card drawn is a spade', F: 'the card drawn is an ace' (ii) E: 'the card drawn is black', F: 'the card drawn is a king' (iii) E: 'the card drawn is a king or queen', F: 'the card drawn is a queen or jack'
(i) P(E)=13/52=1/4, P(F)=4/52=1/13, P(EF)=1/52 (ace of spades)
P(E)*P(F)=(1/4)(1/13)=1/52=P(EF) ✓ Independent.
(ii) P(E)=26/52=1/2, P(F)=4/52=1/13, P(EF)=2/52=1/26 (black kings)
P(E)*P(F)=(1/2)(1/13)=1/26=P(EF) ✓ Independent.
(iii) P(E)=8/52=2/13, P(F)=8/52=2/13, P(EF)=4/52=1/13 (queens: in both E and F)
P(E)*P(F)=(2/13)(2/13)=4/169 ≠ 1/13=13/169
Not independent.
Q16In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English newspapers. (b) If she reads Hindi newspaper, find the probability that she reads English newspaper. (c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
P(H) = 0.6, P(E) = 0.4, P(H intersection E) = 0.2
(a) P(H' intersection E') = 1 - P(H union E)
= 1 - [P(H) + P(E) - P(H intersection E)]
= 1 - [0.6 + 0.4 - 0.2] = 1 - 0.8 = 0.2
(b) P(E|H) = P(H intersection E)/P(H) = 0.2/0.6 = 1/3
(c) P(H|E) = P(H intersection E)/P(E) = 0.2/0.4 = 1/2
Q17The probability of obtaining an even prime number on each die, when a pair of dice is rolled, is (A) 0 (B) 1/3 (C) 1/12 (D) 1/36
The only even prime is 2.
P(2 on first die AND 2 on second die) = (1/6)(1/6) = 1/36
Answer: (D) 1/36
Q18Two events A and B will be independent if (A) A and B are mutually exclusive (B) P(A'B') = [1-P(A)][1-P(B)] (C) P(A) = P(B) (D) P(A)+P(B) = 1
This is the same as Q8. For independent events:
P(A' intersection B') = [1-P(A)][1-P(B)] ✓
Answer: (B)
13.3Exercise 13.3 - Bayes' Theorem
Q1An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Case 1: 1st ball is red (prob 5/10 = 1/2). Now urn has 7R + 5B = 12 balls. P(2nd red) = 7/12.
Case 2: 1st ball is black (prob 5/10 = 1/2). Now urn has 5R + 7B = 12 balls. P(2nd red) = 5/12.
Total P(2nd red) = (1/2)(7/12) + (1/2)(5/12) = 7/24 + 5/24 = 12/24 = 1/2
Q2A bag contains 4 red and 4 black balls. Another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag. If the ball drawn is red, find the probability that it is drawn from the first bag.
P(Bag I) = P(Bag II) = 1/2
P(Red|Bag I) = 4/8 = 1/2
P(Red|Bag II) = 2/8 = 1/4
By Bayes' theorem:
P(Bag I|Red) = P(Red|Bag I)*P(Bag I) / [P(Red|Bag I)*P(Bag I) + P(Red|Bag II)*P(Bag II)]
= (1/2)(1/2) / [(1/2)(1/2) + (1/4)(1/2)]
= (1/4) / (1/4 + 1/8)
= (1/4) / (3/8)
= (1/4)(8/3) = 2/3
Q3Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hosteller?
P(H) = 0.6, P(D) = 0.4
P(A|H) = 0.3, P(A|D) = 0.2
P(A) = P(A|H)P(H) + P(A|D)P(D) = 0.18 + 0.08 = 0.26
P(H|A) = P(A|H)P(H)/P(A) = 0.18/0.26 = 18/26 = 9/13
Q4In 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?
P(K) = 3/4, P(G) = 1/4
P(Correct|K) = 1, P(Correct|G) = 1/4
P(Correct) = P(C|K)P(K) + P(C|G)P(G) = 1(3/4) + (1/4)(1/4) = 3/4 + 1/16 = 13/16
P(K|Correct) = P(C|K)P(K)/P(C) = (3/4)/(13/16) = (3/4)(16/13) = 12/13
Q5A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
P(D) = 0.001, P(D') = 0.999
P(+|D) = 0.99, P(+|D') = 0.005
P(+) = P(+|D)P(D) + P(+|D')P(D')
= 0.99(0.001) + 0.005(0.999)
= 0.00099 + 0.004995 = 0.005985
P(D|+) = P(+|D)P(D)/P(+) = 0.00099/0.005985
= 990/5985 = 22/133 ≈ 0.1654
So even with a positive test, there is only about a 16.5% chance the person actually has the disease. This is because the disease is very rare (0.1%).
Q6There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
P(C1) = P(C2) = P(C3) = 1/3
P(H|C1) = 1 (two-headed), P(H|C2) = 3/4 (biased), P(H|C3) = 1/2 (fair)
P(H) = (1)(1/3) + (3/4)(1/3) + (1/2)(1/3) = 1/3 + 1/4 + 1/6 = 4/12 + 3/12 + 2/12 = 9/12 = 3/4
P(C1|H) = P(H|C1)*P(C1)/P(H) = (1)(1/3)/(3/4) = (1/3)(4/3) = 4/9
Q7An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
P(S) = 2000/12000 = 1/6, P(C) = 4000/12000 = 1/3, P(T) = 6000/12000 = 1/2
P(A|S) = 0.01, P(A|C) = 0.03, P(A|T) = 0.15
P(A) = 0.01(1/6) + 0.03(1/3) + 0.15(1/2)
= 1/600 + 1/100 + 3/40
= 1/600 + 6/600 + 45/600 = 52/600 = 13/150
P(S|A) = P(A|S)P(S)/P(A) = (0.01)(1/6)/(13/150)
= (1/600)/(13/150) = (1/600)(150/13) = 150/7800 = 1/52
Q8A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine A?
P(A) = 0.6, P(B) = 0.4
P(D|A) = 0.02, P(D|B) = 0.01
P(D) = 0.02(0.6) + 0.01(0.4) = 0.012 + 0.004 = 0.016
P(A|D) = P(D|A)P(A)/P(D) = 0.012/0.016 = 12/16 = 3/4
Q9Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product is introduced by the second group.
P(G1) = 0.6, P(G2) = 0.4
P(N|G1) = 0.7, P(N|G2) = 0.3
P(N) = 0.7(0.6) + 0.3(0.4) = 0.42 + 0.12 = 0.54
P(G2|N) = P(N|G2)P(G2)/P(N) = 0.12/0.54 = 12/54 = 2/9
Q10Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
P(D1) = P(5 or 6) = 2/6 = 1/3
P(D2) = P(1,2,3,4) = 4/6 = 2/3
P(1 head | D1) = P(exactly 1 head in 3 tosses) = C(3,1)/2^3 = 3/8
P(1 head | D2) = P(head in 1 toss) = 1/2
P(1 head) = (3/8)(1/3) + (1/2)(2/3) = 1/8 + 1/3 = 3/24 + 8/24 = 11/24
P(D2 | 1 head) = P(1H|D2)P(D2)/P(1H) = (1/3)/(11/24) = (1/3)(24/11) = 8/11
Q11A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
P(A) = 0.5, P(B) = 0.3, P(C) = 0.2
P(D|A) = 0.01, P(D|B) = 0.05, P(D|C) = 0.07
P(D) = 0.01(0.5) + 0.05(0.3) + 0.07(0.2)
= 0.005 + 0.015 + 0.014 = 0.034
P(A|D) = 0.005/0.034 = 5/34 = 5/34
Q12A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
P(lost is diamond) = 13/52 = 1/4
P(lost is not diamond) = 39/52 = 3/4
P(2 diamonds from 51 | lost is diamond): 12 diamonds remain from 51 cards.
P = C(12,2)/C(51,2) = 66/1275
P(2 diamonds from 51 | lost is not diamond): 13 diamonds remain from 51 cards.
P = C(13,2)/C(51,2) = 78/1275
By Bayes':
P(diamond lost | 2 drawn are diamonds)
= [66/1275 * 1/4] / [66/1275 * 1/4 + 78/1275 * 3/4]
= [66/4] / [66/4 + 234/4]
= 66/300 = 11/50
Q13Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was a head is (A) 4/5 (B) 1/2 (C) 1/5 (D) 2/5
P(truth) = 4/5, P(lies) = 1/5
P(H) = P(T) = 1/2
P(reports H | H) = 4/5 (tells truth)
P(reports H | T) = 1/5 (lies)
P(reports H) = (4/5)(1/2) + (1/5)(1/2) = 4/10 + 1/10 = 5/10 = 1/2
P(H | reports H) = [(4/5)(1/2)]/(1/2) = (2/5)/(1/2) = 4/5
Answer: (A) 4/5
Q14If A and B are two events such that A is a subset of B and P(B) != 0, then which of the following is correct? (A) P(A|B) = P(B)/P(A) (B) P(A|B) < P(A) (C) P(A|B) >= P(A) (D) None of these
Since A is a subset of B: A intersection B = A.
P(A|B) = P(A intersection B)/P(B) = P(A)/P(B)
Since A subset of B: P(A) <= P(B), so P(B) >= P(A), so P(A)/P(B) <= 1.
Also P(A)/P(B) >= P(A) (since P(B) <= 1, dividing by a number <= 1 gives a result >= the original).
So P(A|B) = P(A)/P(B) >= P(A) (since 0 < P(B) <= 1).
Answer: (C) P(A|B) >= P(A)
MiscMiscellaneous Exercise on Chapter 13
Q1A and B are two events such that P(A) != 0. Find P(B|A), if (i) A is a subset of B, (ii) A intersection B is empty.
(i) A subset of B means A intersection B = A.
P(B|A) = P(A intersection B)/P(A) = P(A)/P(A) = 1
(ii) A intersection B = empty.
P(B|A) = P(empty)/P(A) = 0/P(A) = 0
Q2A couple has two children. Find the probability that both children are males, if it is known that at least one of the children is male.
S = {MM, MF, FM, FF}
A = both males = {MM}
B = at least one male = {MM, MF, FM}
P(A|B) = P(A intersection B)/P(B) = (1/4)/(3/4) = 1/3
Q3Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.
P(M) = P(F) = 1/2
P(G|M) = 0.05, P(G|F) = 0.0025
P(G) = 0.05(1/2) + 0.0025(1/2) = 0.025 + 0.00125 = 0.02625
P(M|G) = 0.025/0.02625 = 25/26.25 = 2500/2625 = 100/105 = 20/21
Q4Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 are right-handed? [Use Bernoulli trials if needed, or combination formula]
This involves Binomial distribution: n=10, p=0.9, find P(X<=6).
P(X <= 6) = 1 - P(X >= 7) = 1 - [P(7)+P(8)+P(9)+P(10)]
P(X=k) = C(10,k)(0.9)^k(0.1)^(10-k)
P(10) = (0.9)^10 = 0.3487
P(9) = 10(0.9)^9(0.1) = 10(0.3874)(0.1) = 0.3874
P(8) = 45(0.9)^8(0.01) = 45(0.4305)(0.01) = 0.1937
P(7) = 120(0.9)^7(0.001) = 120(0.4783)(0.001) = 0.0574
P(X >= 7) = 0.3487 + 0.3874 + 0.1937 + 0.0574 = 0.9872
P(X <= 6) = 1 - 0.9872 = 0.0128
Note: This question uses Bernoulli trials which may be outside the rationalised syllabus. The answer is approximately 0.0128.
Q5An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y'. A ball is drawn at random from the urn, its mark is noted and it is replaced. If 6 balls are drawn in this way, find the probability that (i) all will bear 'X' mark (ii) not more than 2 will bear 'Y' mark (iii) at least one ball will bear 'Y' mark.
p = P(X) = 10/25 = 2/5, q = P(Y) = 15/25 = 3/5, n = 6
(i) P(all X) = (2/5)^6 = 64/15625 = 64/15625
(ii) P(Y <= 2) = P(Y=0) + P(Y=1) + P(Y=2)
= C(6,0)(3/5)^0(2/5)^6 + C(6,1)(3/5)^1(2/5)^5 + C(6,2)(3/5)^2(2/5)^4
= 64/15625 + 6(3/5)(32/3125) + 15(9/25)(16/625)
= 64/15625 + 576/15625 + 2160/15625
= 2800/15625 = 112/625
(iii) P(at least one Y) = 1 - P(no Y) = 1 - P(all X)
= 1 - 64/15625 = 15561/15625
Q6In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6. What is the probability that he will knock down fewer than 2 hurdles?
n=10, p(clear)=5/6, p(knock)=1/6. Let X = hurdles knocked down.
P(X < 2) = P(X=0) + P(X=1)
= C(10,0)(1/6)^0(5/6)^10 + C(10,1)(1/6)^1(5/6)^9
= (5/6)^10 + 10(1/6)(5/6)^9
= (5/6)^9[(5/6) + 10/6]
= (5/6)^9 * (5/6 + 5/3)
= (5/6)^9 * (5/6 + 10/6)
= (5/6)^9 * (15/6)
= (5/6)^9 * (5/2)
= (5/2)(5/6)^9
Numerically: (5/2)(5^9/6^9) = 5^10/(2*6^9) = 9765625/20155392 ≈ 0.4845
Q7A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.
We need exactly 2 sixes in the first 5 throws AND a six on the 6th throw.
P(2 sixes in 5 throws) = C(5,2)(1/6)^2(5/6)^3
= 10 * (1/36) * (125/216) = 1250/7776
P(six on 6th throw) = 1/6
P(3rd six on 6th throw) = C(5,2)(1/6)^2(5/6)^3 * (1/6)
= 1250/7776 * 1/6 = 1250/46656
= 625/23328
Q8If a leap year is selected at random, what is the probability that it will contain 53 Tuesdays?
A leap year has 366 days = 52 weeks + 2 extra days.
So there are definitely 52 Tuesdays. For 53 Tuesdays, at least one of the 2 extra days must be Tuesday.
The 2 extra days can be: {Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun}
7 equally likely cases.
53 Tuesdays occur if the extra days include Tuesday: {Mon-Tue, Tue-Wed} → 2 cases.
P(53 Tuesdays) = 2/7
Q9An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes.
P(success) = 2/3, P(failure) = 1/3, n = 6
P(X >= 4) = P(4) + P(5) + P(6)
P(4) = C(6,4)(2/3)^4(1/3)^2 = 15 * 16/81 * 1/9 = 15*16/729 = 240/729
P(5) = C(6,5)(2/3)^5(1/3)^1 = 6 * 32/243 * 1/3 = 192/729
P(6) = C(6,6)(2/3)^6 = 64/729
P(X >= 4) = (240 + 192 + 64)/729 = 496/729 = 496/729
Q10How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?
P(at least one head) = 1 - P(all tails) = 1 - (1/2)^n > 0.9
(1/2)^n < 0.1
n * log(1/2) < log(0.1)
-n * 0.3010 < -1
n > 1/0.3010 = 3.32
So n >= 4.
The man must toss the coin at least 4 times.
Verification: 1 - (1/2)^4 = 1 - 1/16 = 15/16 = 0.9375 > 0.9 ✓
1 - (1/2)^3 = 7/8 = 0.875 < 0.9 ✗
Stuck on a question? AI explains any step in detail - in English or Hindi.
Ask AI TutorPrevious Year Board Exam Questions
A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is noted and this ball along with 2 additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, find the probability that the ball drawn is red.
Case 1: 1st red (4/10). Bag: 6R+6B=12. P(2nd red)=6/12=1/2.
Case 2: 1st black (6/10). Bag: 4R+8B=12. P(2nd red)=4/12=1/3.
P(red) = (4/10)(1/2)+(6/10)(1/3) = 2/10+2/10 = 2/5
In a class, 5% of boys and 10% of girls have an IQ of more than 150. In the class, 60% are boys and rest are girls. If a student is selected at random and found to have IQ > 150, find the probability that the student is a boy.
P(B)=0.6, P(G)=0.4. P(IQ>150|B)=0.05, P(IQ>150|G)=0.10.
P(IQ>150) = 0.05(0.6)+0.10(0.4) = 0.03+0.04 = 0.07
P(B|IQ>150) = 0.03/0.07 = 3/7
A and B independently try to solve a problem. The probability that A solves it is 1/2 and B solves it is 1/3. Find the probability that: (i) both solve it, (ii) the problem is solved.
(i) P(AB) = (1/2)(1/3) = 1/6
(ii) P(solved) = 1-P(neither) = 1-(1/2)(2/3) = 1-1/3 = 2/3
If P(A) = 0.4, P(B) = 0.3 and P(A union B) = 0.6, find P(A|B) and P(B|A).
P(AB) = 0.4+0.3-0.6 = 0.1
P(A|B) = 0.1/0.3 = 1/3
P(B|A) = 0.1/0.4 = 1/4
A bag contains 5 red and 3 black balls. Two balls are drawn at random without replacement. Find the probability that both are red.
P = (5/8)(4/7) = 20/56 = 5/14
There are two bags. Bag I has 3 red and 4 black balls, Bag II has 5 red and 6 black balls. One bag is chosen at random and a ball is drawn. If the ball is red, find the probability that it came from Bag I.
P(I)=P(II)=1/2. P(R|I)=3/7, P(R|II)=5/11.
P(R)=(3/7)(1/2)+(5/11)(1/2) = 3/14+5/22 = 33/154+35/154 = 68/154 = 34/77
P(I|R) = (3/14)/(34/77) = (3/14)(77/34) = 231/476 = 33/68 = 33/68
If A and B are independent events, P(A)=0.3, P(B)=0.4. Find P(A intersection B), P(A union B), P(A|B), P(B|A).
P(AB) = 0.12. P(AuB) = 0.58.
P(A|B) = 0.12/0.4 = 0.3 = P(A) (as expected for independent events).
P(B|A) = 0.12/0.3 = 0.4 = P(B). ✓
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.
P(truth)=3/4, P(lies)=1/4. P(six)=1/6, P(not six)=5/6.
P(reports 6|six)=3/4, P(reports 6|not six)=1/4.
P(reports 6) = (3/4)(1/6)+(1/4)(5/6) = 3/24+5/24 = 8/24 = 1/3.
P(six|reports 6) = (3/24)/(1/3) = (1/8)(3) = 3/8
A factory has 3 machines A, B, C producing 50%, 30%, 20% of total output. Defective rates: A=3%, B=4%, C=5%. An item is selected and found defective. Find P(from C).
P(D) = 0.03(0.5)+0.04(0.3)+0.05(0.2) = 0.015+0.012+0.01 = 0.037.
P(C|D) = 0.01/0.037 = 10/37 = 10/37
Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of aces.
P(ace) = 4/52 = 1/13, P(not ace) = 12/13. With replacement, n=2.
P(X=0) = (12/13)^2 = 144/169
P(X=1) = 2(1/13)(12/13) = 24/169
P(X=2) = (1/13)^2 = 1/169
Mean = 0(144/169)+1(24/169)+2(1/169) = 26/169 = 2/13
If P(A) = 0.6, P(B) = 0.3, P(A intersection B) = 0.2. Find P(A|B).
P(A|B) = P(A intersection B)/P(B) = 0.2/0.3 = 2/3
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): If P(A) = 0.4 and P(A intersection B) = 0.2, then P(B|A) = 0.5.
Reason (R): P(B|A) = P(A intersection B)/P(A).
Answer: (A)
P(B|A) = 0.2/0.4 = 0.5. Both true, R explains A.
AR2
Assertion (A): If A and B are independent, then A and B' are also independent.
Reason (R): P(A intersection B') = P(A) - P(A intersection B) = P(A) - P(A)P(B) = P(A)(1-P(B)) = P(A)P(B').
Answer: (A)
The algebraic proof in the reason is correct and proves the assertion.
AR3
Assertion (A): If A and B are mutually exclusive and P(A) > 0, P(B) > 0, then A and B cannot be independent.
Reason (R): For mutually exclusive events, P(A intersection B) = 0, but for independent events P(A intersection B) = P(A)P(B) > 0.
Answer: (A)
P(AB)=0 but P(A)P(B)>0 means they can't be equal. Both true, R explains A.
AR4
Assertion (A): P(A|B) + P(A'|B) = 1.
Reason (R): A and A' form a partition of the sample space.
Answer: (A)
P(A|B)+P(A'|B) = P(AB)/P(B)+P(A'B)/P(B) = [P(AB)+P(A'B)]/P(B) = P(B)/P(B) = 1.
AR5
Assertion (A): Bayes' theorem is used to find the probability of a cause given the effect.
Reason (R): Bayes' theorem reverses the conditioning - from P(effect|cause) to P(cause|effect).
Answer: (A)
Both true. Bayes' theorem updates prior probabilities using observed evidence.
Case Study Questions
A certain disease affects 1 in 1000 people. A diagnostic test has a sensitivity of 99% (correctly identifies 99% of sick people) and a specificity of 95% (correctly identifies 95% of healthy people). A person tests positive.
(a) What is the probability that a randomly selected person has the disease? (1 mark)
P(D) = 1/1000 = 0.001
(b) What is the probability that a healthy person tests positive (false positive)? (1 mark)
False positive rate = 1 - specificity = 1 - 0.95 = 0.05 (5%)
(c) Using Bayes' theorem, find the probability that a person who tests positive actually has the disease. (1 mark)
P(+|D)=0.99, P(+|D')=0.05, P(D)=0.001, P(D')=0.999
P(+) = 0.99(0.001)+0.05(0.999) = 0.00099+0.04995 = 0.05094
P(D|+) = 0.00099/0.05094 = 0.0194 (about 1.94%)
(d) Why is the probability of actually having the disease so low despite a positive test? (1 mark)
Because the disease is very rare (0.1%). Even though the test is accurate, the large number of healthy people tested generates many false positives, which overwhelm the few true positives. This is the base rate fallacy.
A factory has three production lines A, B, and C producing 50%, 30%, and 20% of total output respectively. The defect rates are: A produces 2% defective, B produces 3% defective, and C produces 5% defective items.
(a) What is the probability that a randomly selected item is defective? (1 mark)
P(D) = 0.02(0.5)+0.03(0.3)+0.05(0.2) = 0.01+0.009+0.01 = 0.029 (2.9%)
(b) If an item is found defective, what is the probability it came from line A? (1 mark)
P(A|D) = P(D|A)P(A)/P(D) = 0.01/0.029 = 10/29
(c) If an item is found defective, which production line is it most likely from? (1 mark)
P(A|D)=10/29, P(B|D)=0.009/0.029=9/29, P(C|D)=0.01/0.029=10/29.
Lines A and C are equally likely at 10/29 each, with B slightly lower at 9/29.
(d) What is the probability that a non-defective item came from line B? (1 mark)
P(B|D')=P(D'|B)P(B)/P(D')=0.97(0.3)/0.971=0.291/0.971=291/971
MCQ Practice
1. If P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6, then P(A union B) is:
(A) 0.24
(B) 0.3
(C) 0.48
(D) 0.96
Answer: (D) - P(AB)=P(B|A)*P(A)=0.24. P(AuB)=0.4+0.8-0.24=0.96.
2. If A and B are independent events with P(A)=0.3, P(B)=0.4, then P(A intersection B) is:
(A) 0.12
(B) 0.1
(C) 0.7
(D) 0.58
Answer: (A) - Independent: P(AB)=P(A)*P(B)=0.3*0.4=0.12.
3. If P(A) = 1/2, P(B) = 0, then P(A|B) is:
(A) 0
(B) 1/2
(C) not defined
(D) 1
Answer: (C) - P(A|B)=P(AB)/P(B)=?/0. Division by zero, so not defined.
4. If A and B are such that P(A union B) = 5/6, P(A intersection B) = 1/3, P(B') = 1/2, then P(A) is:
(A) 1/3
(B) 2/3
(C) 1/2
(D) 1/6
Answer: (B) - P(B)=1-1/2=1/2. P(AuB)=P(A)+P(B)-P(AB). 5/6=P(A)+1/2-1/3. P(A)=5/6-1/6=4/6=2/3.
5. A die is thrown. If E is the event 'number is multiple of 3' and F is 'number is even'. Then P(E|F) is:
(A) 1/2
(B) 1/3
(C) 1/6
(D) 2/3
Answer: (B) - E={3,6}, F={2,4,6}. EF={6}. P(E|F)=(1/6)/(3/6)=1/3.
6. P(A intersection B) = P(A)*P(B) is the condition for:
(A) Mutually exclusive events
(B) Dependent events
(C) Independent events
(D) Exhaustive events
Answer: (C) - P(AB)=P(A)*P(B) is the definition of independent events.
7. If P(A) = 3/10, P(B) = 2/5 and P(A union B) = 3/5, then P(B|A) + P(A|B) is:
(A) 1/4
(B) 1/3
(C) 5/12
(D) 7/12
Answer: (D) - P(AB)=3/10+2/5-3/5=3/10+4/10-6/10=1/10. P(B|A)=(1/10)/(3/10)=1/3. P(A|B)=(1/10)/(2/5)=1/4. Sum=1/3+1/4=7/12.
8. A bag has 5 red, 4 blue, 3 green balls. Three balls are drawn. P(all three are green) is:
(A) 1/220
(B) 3/220
(C) 1/22
(D) 3/22
Answer: (A) - C(3,3)/C(12,3)=1/220.
9. If P(A)=0.6, P(B)=0.3, P(A intersection B)=0.2, then P(A|B') is:
(A) 2/7
(B) 4/7
(C) 3/7
(D) 5/7
Answer: (B) - P(AB')=P(A)-P(AB)=0.6-0.2=0.4. P(B')=0.7. P(A|B')=0.4/0.7=4/7.
10. The probability that a student is not a swimmer is 1/5. Then the probability that out of 5 students, 4 are swimmers is:
(A) C(5,4)(4/5)^4(1/5)
(B) (4/5)^4(1/5)
(C) C(5,1)(1/5)(4/5)^4
(D) None
Answer: (A) - P(swimmer)=4/5. P(X=4)=C(5,4)(4/5)^4(1/5)^1=5*(256/3125)*(1/5)=1280/3125=256/625.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
If P(A) = 0.6, P(B) = 0.3, P(A intersection B) = 0.2, find P(A|B).
P(A|B) = 0.2/0.3 = 2/3.
If A and B are independent with P(A)=1/3, P(B)=1/4, find P(A intersection B).
P(AB) = P(A)*P(B) = 1/12.
If P(A) = 1/2 and P(B) = 0, what is P(A|B)?
Not defined (P(B) = 0, cannot divide by 0).
Two cards drawn without replacement from 52 cards. P(both kings)?
(4/52)(3/51) = 12/2652 = 1/221.
A die is thrown. E={1,3,5}, F={2,3}. Are E and F independent?
P(E)=1/2, P(F)=1/3. P(EF)=P({3})=1/6. P(E)*P(F)=1/6=P(EF). Yes, independent.
2-3 Mark Questions
A and B try to solve a problem independently. P(A solves)=1/2, P(B solves)=1/3. Find probability the problem is solved.
P(solved)=1-P(neither)=1-(1/2)(2/3)=1-1/3=2/3
A bag has 4 red, 4 black balls. Another has 2 red, 6 black. One bag chosen at random, one ball drawn. If red, find P(from Bag I).
P(R|I)=1/2, P(R|II)=1/4. P(R)=(1/2)(1/2)+(1/4)(1/2)=3/8.
P(I|R)=(1/4)/(3/8)=2/3
If P(A)=3/5, P(B)=1/5, find P(A union B) if A,B are independent.
P(AB)=(3/5)(1/5)=3/25. P(AuB)=3/5+1/5-3/25=15/25+5/25-3/25=17/25
60% students read Hindi paper, 40% English, 20% both. Find P(neither).
P(HuE)=0.6+0.4-0.2=0.8. P(neither)=1-0.8=0.2
3-4 Mark Questions
In a test, student knows answer with prob 3/4, guesses with prob 1/4. If guessing, P(correct)=1/4. Given correct answer, find P(knew the answer).
P(C|K)=1, P(C|G)=1/4. P(C)=3/4+1/16=13/16.
P(K|C)=(3/4)/(13/16)=12/13
Three machines A,B,C produce 50%,30%,20% of items. Defect rates: 3%,4%,5%. Find P(item from C given defective).
P(D)=0.015+0.012+0.01=0.037.
P(C|D)=0.01/0.037=10/37
A factory has two machines: A produces 60% items (2% defective), B produces 40% (1% defective). Find P(from A given defective).
P(D)=0.012+0.004=0.016.
P(A|D)=0.012/0.016=3/4
Key Concepts
Important Formulas
Study Plan - Probability
Conditional Probability
Learn P(A|B) formula and properties. Solve Ex 13.1 completely (17 questions)
Multiplication Theorem & Independence
Master independent events test. Complete Ex 13.2 (18 questions)
Bayes' Theorem
Learn total probability + Bayes' formula. Solve Ex 13.3 Q1-Q8
Bayes' Applications
Complete Ex 13.3 Q9-Q14. Focus on word problems (medical, factory, card problems)
Miscellaneous & PYQs
Solve Miscellaneous Exercise and past 5 years CBSE questions on probability
Revision and Mock Practice
Practice case study format, MCQs, and mixed problems. Time yourself on 5-mark Bayes' questions
Common Mistakes to Avoid
- ✗Confusing P(A|B) with P(B|A) - these are generally NOT equal
- ✗Forgetting that P(A|B) requires P(B) != 0
- ✗Using multiplication rule for independent events when events are NOT independent
- ✗In Bayes' theorem: forgetting to compute the total probability P(E) in the denominator
- ✗Confusing 'mutually exclusive' with 'independent' - they are different concepts
- ✗In card/ball problems: not distinguishing between 'with replacement' and 'without replacement'
- ✗Arithmetic errors in fractions when computing conditional probabilities
- ✗Not identifying all cases (partitions) when applying total probability theorem
- ✗Assuming events are independent without verification: always check P(AB) = P(A)*P(B)
Tips for Board Exam
- ✓Bayes' theorem is the most frequently tested 5-mark topic - memorize the formula and practice 10+ problems
- ✓For Bayes': always set up a table with P(Ei), P(A|Ei), and P(A|Ei)*P(Ei) for each case
- ✓Quick independence check: P(A intersection B) = P(A)*P(B)? If yes, independent
- ✓P(at least one) = 1 - P(none) is the easiest approach for 'at least' problems
- ✓For conditional probability MCQs: directly apply P(A|B) = P(AB)/P(B)
- ✓Common Bayes' scenarios: factory defects, medical testing, bag selection - practice all three
- ✓Remember: P(A|B) + P(A'|B) = 1 (useful for quick calculations)
- ✓In case study format: read the entire context before computing - identify what each probability represents
- ✓For 'without replacement' problems: use multiplication rule with updated probabilities
Practice Probability with AI Tutor
Stuck on a question? Ask our AI tutor to explain step-by-step in English or Hindi. Choose from 10 learning modes - Lecture, Quiz, Flashcards, and more. Free for 3 chapters per day.
Start Probability - Choose Your ModeAbout Probability - Class 12 Mathematics
Probability is Chapter 13 of NCERT Class 12 Mathematics (Part II) and the final chapter. It builds on Class 11 probability by introducing conditional probability P(A|B) = P(A intersection B)/P(B), the multiplication theorem, and the concept of independent events (where P(A intersection B) = P(A)*P(B)). The chapter culminates in Bayes' theorem, which reverses conditional probabilities - from P(effect|cause) to P(cause|effect). Applications include medical diagnostic accuracy, factory quality control, card and ball drawing problems, and exam prediction. This chapter carries 8 marks as an independent unit in the CBSE board exam, with Bayes' theorem being the most frequently tested 5-mark topic. The 2026-27 rationalised syllabus covers exercises 13.1 (conditional probability), 13.2 (multiplication theorem and independence), 13.3 (Bayes' theorem), and a Miscellaneous Exercise.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 13?
Per the 2026-27 rationalised syllabus, Chapter 13 has 3 exercises plus a Miscellaneous Exercise: Ex 13.1 (17 questions on conditional probability), Ex 13.2 (18 questions on multiplication theorem and independence), Ex 13.3 (14 questions on Bayes' theorem), and Miscellaneous (10 questions).
What is the weightage of Probability in the CBSE board exam?
Probability is an independent unit carrying 8 marks. Typically: 1 MCQ (1 mark) + 1 short answer on conditional probability or independence (2-3 marks) + 1 long answer on Bayes' theorem (5 marks).
What is Bayes' theorem and when is it used?
Bayes' theorem finds P(cause|effect) when P(effect|cause) is known: P(Ei|A) = P(A|Ei)*P(Ei) / sum(P(A|Ej)*P(Ej)). It is used when you observe an outcome and want to determine which cause was most likely (e.g., which factory produced a defective item, whether a patient actually has a disease given a positive test).
What is the difference between independent and mutually exclusive events?
Independent events: occurrence of one does not affect the other (P(AB) = P(A)*P(B)). Mutually exclusive: both cannot occur together (P(AB) = 0). If P(A) > 0 and P(B) > 0, mutually exclusive events are NEVER independent (since P(A)*P(B) > 0 but P(AB) = 0).
Are Bernoulli trials and random variables in the 2026-27 syllabus?
No. The 2026-27 rationalised NCERT textbook removed Exercises 13.4 (Bernoulli Trials and Binomial Distribution) and 13.5 (random variables, mean, variance). Only conditional probability, independence, and Bayes' theorem are in the current syllabus. However, some Miscellaneous questions may still reference binomial concepts.