Probability — Previous Year Questions (Class 12 Mathematics)
Probability (Class 12) covers conditional probability, Bayes' theorem, and probability distributions. It extends Class 10 foundations to complex real-world
TL;DR: Probability (Class 12) covers conditional probability, Bayes' theorem, and probability distributions. It extends Class 10 foundations to complex real-…
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Probability (Class 12) covers conditional probability, Bayes' theorem, and probability distributions. It extends Class 10 foundations to complex real-world
Probability — Previous Year Questions with Solutions
Q (2023, 2 marks): If P(A) = 0.4, P(B) = 0.6, and P(A ∩ B) = 0.2, find P(A | B).
Answer: Conditional probability: P(A | B) = P(A ∩ B) / P(B)
Given: P(A ∩ B) = 0.2, P(B) = 0.6
P(A | B) = 0.2 / 0.6 = 1/3 ≈ 0.333
Q (2022, 2 marks): Two cards are drawn from a standard deck without replacement. Find P(both are aces).
Answer: Method 1: Using conditional probability
P(both aces) = P(1st ace) × P(2nd ace | 1st ace)
= (4/52) × (3/51)
= (1/13) × (3/51)
= 3/663
= 1/221
≈ 0.00452
Method 2: Using combinations
P(both aces) = C(4,2) / C(52,2)
= 6 / 1326
= 1/221
Q (2023, 3 marks): A bag contains 3 red, 4 blue, and 5 green balls. Two balls are drawn with replacement. Find P(both same color).
Answer: Total balls = 3 + 4 + 5 = 12
With replacement, each draw is independent
P(both same) = P(both red) + P(both blue) + P(both green)
P(both red) = (3/12) × (3/12) = (1/4) × (1/4) = 1/16
P(both blue) = (4/12) × (4/12) = (1/3) × (1/3) = 1/9
P(both green) = (5/12) × (5/12) = 25/144
P(both same) = 1/16 + 1/9 + 25/144
Finding common denominator (144):
= 9/144 + 16/144 + 25/144
= 50/144
= 25/72
Q (2021, 3 marks): Using Bayes' theorem: A disease affects 1% of population. Test accuracy: 95% true positive, 90% true negative. Find P(disease | positive test).
Answer: Let D = disease, + = positive test
Given:
P(D) = 0.01, P(D') = 0.99
P(+ | D) = 0.95 (sensitivity)
P(- | D') = 0.90 (specificity)
P(+ | D') = 0.10 (false positive rate)
Using Bayes' theorem:
P(D | +) = P(+ | D) × P(D) / P(+)
First find P(+):
P(+) = P(+ | D) × P(D) + P(+ | D') × P(D')
= 0.95 × 0.01 + 0.10 × 0.99
= 0.0095 + 0.099
= 0.1085
P(D | +) = 0.95 × 0.01 / 0.1085
= 0.0095 / 0.1085
≈ 0.0876 or 8.76%
So despite positive test, only 8.76% probability of actually having disease (due to low prevalence)
Q (2022, 3 marks): In a class of 60 students, 35 study Math, 40 study Science, 15 study both. Find probability that a random student studies neither.
Answer: Using set theory:
Let M = Math, S = Science
|M| = 35, |S| = 40, |M ∩ S| = 15
|M ∪ S| = |M| + |S| - |M ∩ S|
= 35 + 40 - 15
= 60
Students studying at least one subject = 60
Students studying neither = 60 - 60 = 0
P(neither) = 0 / 60 = 0
Note: All students study at least one subject in this case
Q (2023, 2 marks): A fair die is rolled 3 times. Find P(at least one 6).
Answer: Method 1: Complement
P(at least one 6) = 1 - P(no 6)
P(no 6 in one roll) = 5/6
P(no 6 in three rolls) = (5/6)³ = 125/216
P(at least one 6) = 1 - 125/216 = 91/216
≈ 0.421
Frequently Asked Questions
What is the difference between conditional probability and independent events?
Conditional probability P(A|B) = P(A∩B)/P(B) depends on another event. Independent events satisfy P(A∩B) = P(A)×P(B), meaning P(A|B) = P(A). Independence means one event's outcome doesn't affect the other's probability.
How do you apply Bayes' theorem in real-world problems?
Bayes' theorem: P(A|B) = P(B|A)×P(A)/P(B). Use it when you know the reverse conditional probability and need to find the forward one. Common applications: disease testing, spam detection, quality control, legal evidence evaluation.
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