Probability – Advanced Solved Problems

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Probability – Advanced Solved Problems (With Detailed Solutions)

Probability – Advanced Solved Problems

This page contains advanced-level probability problems with complete step-by-step explanations. These questions are commonly asked in competitive exams, data aptitude tests, statistics foundations, and real-world probability reasoning.

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Problem 1: Conditional Probability

A card is drawn from a standard deck of 52 playing cards. What is the probability that the card is a King, given that the card drawn is a face card?

Solution:

First, recall the formula for conditional probability:

P(A | B) = P(A ∩ B) / P(B)

Let:

  • A = Event that the card is a King
  • B = Event that the card is a face card

Number of face cards = Kings + Queens + Jacks = 3 types × 4 suits = 12 face cards

Number of Kings = 4

Since every King is a face card, A ∩ B = 4

P(A | B) = 4 / 12 = 1 / 3

Answer: The required probability is 1/3.

Problem 2: Bayes Theorem

A factory has three machines A, B, and C producing 25%, 35%, and 40% of total output respectively. Their defect rates are 2%, 3%, and 5% respectively. A randomly selected product is found to be defective. What is the probability that it was produced by Machine B?

Solution:

We apply Bayes’ theorem:

P(B | D) = [P(B) × P(D | B)] / P(D)

Given:

  • P(A) = 0.25, P(D | A) = 0.02
  • P(B) = 0.35, P(D | B) = 0.03
  • P(C) = 0.40, P(D | C) = 0.05

First calculate P(D):

P(D) = (0.25×0.02) + (0.35×0.03) + (0.40×0.05)
P(D) = 0.005 + 0.0105 + 0.020 = 0.0355

Now apply Bayes formula:

P(B | D) = (0.35 × 0.03) / 0.0355 = 0.0105 / 0.0355 ≈ 0.296

Answer: Probability ≈ 0.296 (29.6%)

Problem 3: Independent Events

Two dice are thrown simultaneously. Find the probability that the sum of the numbers is 8, given that the number on the first die is even.

Solution:

First list even numbers on first die: {2, 4, 6}

Possible outcomes when first die is even = 3 × 6 = 18

Now count outcomes where sum = 8 and first die is even:

  • (2,6)
  • (4,4)
  • (6,2)
Total = 3 outcomes

P = 3 / 18 = 1 / 6

Answer: Probability = 1/6

Problem 4: Combination-Based Probability

A committee of 3 is to be formed from 5 men and 4 women. What is the probability that the committee has exactly 2 women?

Solution:

Total people = 9 Total committees = C(9,3) = 84

Favorable cases:

  • Select 2 women from 4 → C(4,2) = 6
  • Select 1 man from 5 → C(5,1) = 5

Total favorable = 6 × 5 = 30

P = 30 / 84 = 5 / 14

Answer: Probability = 5/14

Problem 5: Probability with Replacement

A bag contains 4 red and 6 blue balls. Two balls are drawn one after another with replacement. What is the probability that both balls are red?

Solution:

Total balls = 10

Probability of red in one draw = 4/10 = 2/5

Since replacement occurs, events are independent:

P = (2/5) × (2/5) = 4/25

Answer: Probability = 4/25

Problem 6: Mutually Exclusive Events

What is the probability that a randomly selected number from 1 to 20 is divisible by 3 or 5?

Solution:

Numbers divisible by 3: 3,6,9,12,15,18 → 6 numbers Numbers divisible by 5: 5,10,15,20 → 4 numbers Common number: 15

Using inclusion-exclusion:

Total favorable = 6 + 4 − 1 = 9
P = 9 / 20

Answer: Probability = 9/20

Problem 7: Real-Life Probability Reasoning

The probability that a student passes Mathematics is 0.7 and Science is 0.6. If the probability that the student passes both subjects is 0.5, find the probability that the student passes at least one subject.

Solution:

Use the formula:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P = 0.7 + 0.6 − 0.5 = 0.8

Answer: Probability = 0.8 (80%)

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Conclusion: These advanced probability problems build a strong foundation for competitive exams, statistics, and analytical reasoning. Regular practice improves logical thinking and real-world decision-making.

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