Probability: CBSE Class 12 Previous Year Questions
20 different questions from Probability (NCERT Chapter 13) asked in CBSE Class 12 Maths board exams 2026.
Pick a mark group to practise, each with answers and step-by-step solutions.
Most asked Probability questions
Assertion (A): In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is 32.
Reason (R): For any two events A and B, P(A∣B)=P(B)P(A∪B)
- (A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
- (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- (C)Assertion (A) is true and Reason (R) is false.
- (D)Assertion (A) is false and Reason (R) is true.
Show answer & solution
Answer: (C) Assertion (A) is true and Reason (R) is false.
- Odd outcomes: {1, 3, 5}. Prime among them: {3, 5}.
- P(prime∣odd)=32, so A is true.
- The correct formula is P(A∣B)=P(B)P(A∩B), not with A∪B, so R is false.
For two events A and B such that P(A)=0 and P(B)=1, P(A′/B′)=
- (A)1−P(A/B)
- (B)1−P(A′/B)
- (C)P(B′)1−P(A∩B)
- (D)P(B′)1−P(A∪B)
Show answer & solution
Answer: (D) P(B′)1−P(A∪B)
- P(A′/B′)=P(B′)P(A′∩B′).
- A′∩B′=(A∪B)′, so P(A′∩B′)=1−P(A∪B).
- P(A′/B′)=P(B′)1−P(A∪B).
If E and F are two independent events such that P(E)=103, P(E∪F)=21, then P(E∣F)−P(F∣E) is equal to :
- (A)72
- (B)353
- (C)701
- (D)71
Show answer & solution
Answer: (C) 701
- For independent events, P(E∪F)=P(E)+P(F)−P(E)P(F).
- 21=103+107P(F), so P(F)=72.
- P(E∣F)−P(F∣E)=P(E)−P(F)=103−72=701.
If 3P(A)=P(B)=53 and P(A∣B)=32, then P(A∪B) is :
- (A)53
- (B)51
- (C)152
- (D)52
Show answer & solution
Answer: (D) 52
- P(B)=53, P(A)=51.
- P(A∩B)=P(A∣B)P(B)=32⋅53=52.
- P(A∪B)=51+53−52=52.
Out of two bags, bag I contains 3 red and 4 white balls and bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3 then a ball is drawn at random from bag I, otherwise a ball is drawn at random from bag II. Find the probability that the ball drawn from one of the bags is a red ball.
Show answer & solution
Answer: 2111
- Let E1: die shows 1 or 2 (bag I), E2: die shows 3, 4, 5 or 6 (bag II). P(E1)=31, P(E2)=32.
- Let R: red ball drawn. P(R∣E1)=73, P(R∣E2)=148=74.
- P(R)=31⋅73+32⋅74=213+218=2111.
The probability of simultaneous occurrence of atleast one of the two events X and Y is a. If the probability that exactly one of the events X, Y occurs is b, prove that P(X′)+P(Y′)=2−2a+b.
Show answer & solution
Answer: Proved.
- Given P(X∪Y)=P(X)+P(Y)−P(X∩Y)=a.
- Exactly one occurs: P(X)+P(Y)−2P(X∩Y)=b.
- Subtracting: P(X∩Y)=a−b, so P(X)+P(Y)=a+(a−b)=2a−b.
- P(X′)+P(Y′)=2−[P(X)+P(Y)]=2−2a+b. Hence proved.
The probability of hitting the target by a trained sniper is three times the probability of not hitting the target on a stormy day due to high wind speed.
The sniper fired two shots on the target on a stormy day when wind speed was very high. Find the probability that
(i) target is hit
(ii) atleast one shot misses the target.
Show answer & solution
Answer: (i) 1615 (ii) 167
- Let p = P(hit). p=3(1−p)⇒p=43, P(miss) =41. Shots are independent.
- (i) Target is hit (at least once) =1−P(both miss)=1−161=1615.
- (ii) At least one shot misses =1−P(both hit)=1−169=167.
Mother, Father and Son line up at random for a family picture. Let events E : Son on one end and F : Father in the middle. Find P(E/F).
Show answer & solution
Answer: P(E/F)=1
- Sample space: 3!=6 equally likely arrangements.
- F = {MFS, SFM}, so P(F)=62=31.
- In both arrangements of F the son is at an end, so E∩F=F and P(E∩F)=31.
- P(E/F)=P(F)P(E∩F)=1.
A survey was conducted on the patients who have undergone knee replacement surgeries.
It was found that, Robotic Knee replacement surgeries have 90% success rate.
On a particular day, robotic surgery was performed on three patients, A, B and C, one after the other. Assuming that the success and failure of each surgery is independent of each other, find the probability that :
(i) exactly one surgery is successful,
(ii) at most two surgeries are successful.
Show answer & solution
Answer: (i) 0.027 (ii) 0.271
- P(success) = 0.9, P(failure) = 0.1 for each surgery, independently.
- (i) Exactly one success: 3×0.9×0.1×0.1=0.027.
- (ii) At most two successes = 1 − P(all three succeed) =1−(0.9)3=1−0.729=0.271.
In a school, the probability of holding a debate competition is 31 and that of a quiz competition is 32. In the two participating teams, A has 4 girls and 6 boys and B has 7 girls and 3 boys. If a debate competition is held, the students are selected from team A and for the quiz competition they are selected from team B. If only two students are to be chosen from the teams, then find the probability that one will be a girl and the other a boy.
Show answer & solution
Answer: 4522
- Let E1: debate (team A), E2: quiz (team B); P(E1)=31, P(E2)=32. Let G: one girl and one boy chosen.
- P(G∣E1)=10C24×6=4524, P(G∣E2)=10C27×3=4521.
- P(G)=31⋅4524+32⋅4521=13524+42=13566=4522.
Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on
MonoMath CBSE →