Probability: Conceptual Doubts Cleared (10)

easy • 3 min read
Tags Probability

Question

A bag contains 5 red and 7 blue balls. Two balls are drawn one after another, without replacement. Find (a) the probability that both are red, (b) the probability that the second ball is blue, and (c) given that the second ball is blue, the probability that the first was red. CBSE 2024 boards combined with JEE Main pattern (Bayes’ theorem).

Solution — Step by Step

P(first red)=512P(\text{first red}) = \dfrac{5}{12}.

After one red is drawn, there are 4 red and 7 blue balls left, so P(second red∣first red)=411P(\text{second red} \mid \text{first red}) = \dfrac{4}{11}.

P(both red)=512×411=20132=533P(\text{both red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}

The second ball is blue can happen via two paths:

  • First red, second blue: 512×711=35132\dfrac{5}{12} \times \dfrac{7}{11} = \dfrac{35}{132}
  • First blue, second blue: 712×611=42132\dfrac{7}{12} \times \dfrac{6}{11} = \dfrac{42}{132}

P(second blue)=35+42132=77132=712P(\text{second blue}) = \frac{35 + 42}{132} = \frac{77}{132} = \frac{7}{12}

We need P(first red∣second blue)P(\text{first red} \mid \text{second blue}). By definition:

P(first red∣second blue)=P(first red AND second blue)P(second blue)P(\text{first red} \mid \text{second blue}) = \frac{P(\text{first red AND second blue})}{P(\text{second blue})}

P(first red∣second blue)=35/13277/132=3577=511P(\text{first red} \mid \text{second blue}) = \frac{35/132}{77/132} = \frac{35}{77} = \frac{5}{11}

Final answers: (a) 533\dfrac{5}{33}, (b) 712\dfrac{7}{12}, (c) 511\dfrac{5}{11}.

Why This Works

The interesting result is part (b): the probability that the second ball is blue is 712\dfrac{7}{12} — the same as the probability the first ball is blue. This isn’t a coincidence. By symmetry, every ball in the bag has the same probability of being any given draw. The “second draw” is no more special than the first.

Part (c) is Bayes’ theorem in disguise: we know the outcome of the second draw and want the probability of the first. The formula P(A∣B)=P(A∩B)/P(B)P(A|B) = P(A \cap B)/P(B) handles it directly.

Alternative Method

For part (b), use the symmetry argument directly: any specific ball is equally likely to be drawn at any specific position, so P(second blue)=P(any single ball blue)=7/12P(\text{second blue}) = P(\text{any single ball blue}) = 7/12.

This is the elegant 5-second answer that toppers use. Students who don’t see the symmetry compute it the long way (Step 2 above).

The biggest error is treating “without replacement” as if it were “with replacement”. With replacement, the probability of both red would be (5/12)2=25/144(5/12)^2 = 25/144. Without replacement, after removing one red, only 4 reds remain — denominator drops to 11.

Whenever you see “without replacement” and “second draw”, look for a Bayes’ theorem opportunity. CBSE 2-mark questions in the conditional probability section follow this pattern almost every year.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

Bayes Theorem — Medical Test Problem Worked Out

hard

Bernoulli trials — probability of exactly 3 successes in 5 trials

medium

Bernoulli Trials — Probability of Exactly 2 Successes in 5 Trials

medium

Birthday problem — probability that 2 of 23 people share a birthday

hard

Conditional Probability — Drawing Cards Example

medium