Probability: PYQ Walkthrough (12)

hard 2 min read

Question

A bag contains 44 red and 66 blue balls. Two balls are drawn at random one after another without replacement. Find the probability that both are red. (JEE Main 2023 pattern)

Solution — Step by Step

Without replacement means the two draws are not independent — the result of the first changes the composition of the bag for the second.

P(R1)=410=25P(R_1) = \frac{4}{10} = \frac{2}{5}

After removing one red, the bag has 33 red and 66 blue, total 99:

P(R2R1)=39=13P(R_2 \mid R_1) = \frac{3}{9} = \frac{1}{3}
P(R1R2)=P(R1)P(R2R1)=2513=215P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 \mid R_1) = \frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15}

Final answer: P=215P = \tfrac{2}{15}.

Why This Works

The chain rule for conditional probability: P(AB)=P(A)P(BA)P(A \cap B) = P(A) P(B|A). Whenever the question says “without replacement”, “given that…”, or “in succession”, reach for conditional probability.

The denominator changes between the two draws because the universe shrinks — one ball is gone.

Alternative Method

Use combinations: number of ways to choose 2 reds from 4 is (42)=6\binom{4}{2} = 6; total ways to choose 2 from 10 is (102)=45\binom{10}{2} = 45. Probability =6/45=2/15= 6/45 = 2/15. Same answer in one line — preferred for MCQ speed.

For “at least one red” or “exactly one red” type variants, switch to combinations directly. Faster than tree diagrams.

Common Mistake

Treating the draws as independent and computing 410410=425\tfrac{4}{10} \cdot \tfrac{4}{10} = \tfrac{4}{25}. This would be correct only with replacement. JEE Main loves this trap — read the question twice.

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