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Official PYQ · 2024Set 1 · Q53Conditional Probability and Bayes TheoremNAT2 marks

GATE CS 2024 Set 1 Q53 — Conditional Probability and Bayes Theorem

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A bag contains 1010 red balls and 1515 blue balls. Two balls are drawn randomly without replacement. Given that the first ball drawn is red, the probability (rounded off to 33 decimal places) that both balls drawn are red is ___________.

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Answer: 0.375

Let us define the following events:

  • R1R_1: The first ball drawn is red.
  • R2R_2: The second ball drawn is red.

We need to find the conditional probability that both balls drawn are red, given that the first ball drawn is red:

P(Both are red∣R1)=P(R1∩R2∣R1)P(\text{Both are red} \mid R_1) = P(R_1 \cap R_2 \mid R_1)

Using the definition of conditional probability:

P(R1∩R2∣R1)=P(R1∩R2)P(R1)=P(R2∣R1)P(R_1 \cap R_2 \mid R_1) = \frac{P(R_1 \cap R_2)}{P(R_1)} = P(R_2 \mid R_1)

Initially, the bag contains:

  • Red balls = 1010
  • Blue balls = 1515
  • Total balls = 2525

Given that the first ball drawn is red (event R1R_1 has occurred) and the drawing is without replacement, the composition of the bag for the second draw becomes:

  • Remaining red balls = 10−1=910 - 1 = 9
  • Remaining blue balls = 1515
  • Total remaining balls = 2424

The probability that the second ball is red, given the first was red, is:

P(R2∣R1)=Number of remaining red ballsTotal remaining balls=924=38=0.375P(R_2 \mid R_1) = \frac{\text{Number of remaining red balls}}{\text{Total remaining balls}} = \frac{9}{24} = \frac{3}{8} = 0.375

Thus, the required probability is 0.375.

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