Visual SolutionPYQ 2024 · Jan Shift 2Standard

Question

Three urns contain 2 white and 3 black, 3 white and 2 black, and 4 white and 1 black balls respectively. One ball is drawn from each urn. The probability of drawing exactly 2 white balls is:

(A)

\frac{3}{5}

(B)

\frac{2}{5}

(C)

\frac{3}{715}

(D)

\frac{19}{50}

Solution Path

3 cases:

(W,W,B)=6/125

(W,B,W)=16/125

(B,W,W)=36/125

. Total

= 58/125

01Question Setup

1/4

Three urns:

2W+3B

3W+2B

4W+1B

. Draw one ball from each. Find

P(\text{exactly 2 white})

Find

P(\text{exactly 2 white})

02Three Cases

2/4

Exactly 2 white means 3 cases: (W,W,B), (W,B,W), (B,W,W). List each probability product.

P(W_i) = \tfrac{2}{5}, \tfrac{3}{5}, \tfrac{4}{5}

03Add CasesKEY INSIGHT

3/4

Case 1:

\tfrac{6}{125}

. Case 2:

\tfrac{16}{125}

. Case 3:

\tfrac{36}{125}

. Sum

= \tfrac{58}{125}

P = \dfrac{6 + 16 + 36}{125} = \dfrac{58}{125}

04Final Answer

4/4

Sum of all three mutually exclusive cases gives the answer.

P = \boxed{\dfrac{58}{125}}

- Answer (A)

Concepts from this question1 concepts unlocked

★

Multiplication Rule

EASY

P(A ∩ B) = P(A) × P(B|A). For sequential events, multiply the probability of each stage given all previous outcomes.

P(A \cap B) = P(A) \cdot P(B \mid A)

The backbone of without-replacement and multi-stage probability. Appears in nearly every JEE probability question involving sequential draws.

Without replacementSequential drawsMulti-stage experiments

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