Visual SolutionPYQ 2024 · Jan Shift 1Easy

Question

A bag contains 4 white and 6 black balls. Two balls are drawn at random one after the other without replacement. The probability that the first drawn ball is white and the second drawn ball is black is:

(A)

\frac{4}{15}

(B)

\frac{2}{7}

(C)

\frac{1}{3}

(D)

\frac{4}{9}

Solution Path

Without replacement means the sample space shrinks after each draw. Multiplication rule:

P(W_1 \cap B_2) = P(W_1) \times P(B_2|W_1) = \frac{4}{10} \cdot \frac{6}{9} = \frac{4}{15}

01Question Setup

1/4

4 white + 6 black = 10 balls. Draw 2 without replacement. Find P(1st White ∩ 2nd Black).

Total:

10

balls

02First Draw - White Ball

2/4

Pick a white ball from 10 total. After removing it, 9 balls remain (3 white + 6 black).

P(W_1) = \dfrac{4}{10} = \dfrac{2}{5}

03Second Draw - Without ReplacementKEY INSIGHT

3/4

With one white removed, 6 black out of 9 remain. Multiply sequential probabilities using the multiplication rule.

P(W_1 \cap B_2) = \dfrac{4}{10} \times \dfrac{6}{9} = \dfrac{4}{15}

04Final Answer

4/4

24/90 simplifies to 4/15. Answer is option (A).

P = \dfrac{4}{15}

- Answer (A)

Concepts from this question3 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

Practice (19 Qs) →

★

Conditional Probability

EASY

P(B|A) = P(A ∩ B) / P(A). The probability of B given that A has already occurred.

P(B \mid A) = \frac{P(A \cap B)}{P(A)}

Without-replacement problems are conditional by nature -the sample space changes after each draw. Misidentifying the reduced sample space is the #1 error.

Without replacementBayes theoremIndependent events test

Practice (10 Qs) →

Without Replacement -Shrinking Sample Space

EASY

After each draw, both the favorable and total counts decrease. The denominator in each successive probability is one less than before.

P(\text{2nd stage}) = \frac{\text{remaining favorable}}{n - 1}

Students often forget to reduce the total. This single concept prevents the most common probability mistake in JEE.

Ball-draw problemsCard problemsSequential selection

Practice (9 Qs) →

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