GATE CS 2024 Set 1 — Question 63

Total number of red balls = 10
Total number of blue balls = 15
Total number of balls = 10 + 15 = 25Let $R_1$ be the event that the first ball drawn is red, and $R_2$ be the event that the second ball drawn is red.
The question asks for the probability that both balls are red given that the first ball is red. This is expressed as the conditional probability $P(R_1 \cap R_2 \mid R_1)$.By the definition of conditional probability:$$P(R_1 \cap R_2 \mid R_1) = \frac{P((R_1 \cap R_2) \cap R_1)}{P(R_1)} = \frac{P(R_1 \cap R_2)}{P(R_1)}$$Using the multiplication rule, $P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 \mid R_1)$. Substituting this back:$$\frac{P(R_1) \cdot P(R_2 \mid R_1)}{P(R_1)} = P(R_2 \mid R_1)$$Since the balls are drawn without replacement, if the first ball is red, there are $10 - 1 = 9$ red balls remaining out of a total of $25 - 1 = 24$ balls.$$P(R_2 \mid R_1) = \frac{9}{24} = \frac{3}{8} = 0.375$$Thus, the probability is 0.375.