A bag contains 10 red balls and 15 blue balls. Two balls are drawn randomly without replacement. Given that the first ball drawn is red, the probability (rounded off to 3 decimal places) that both balls drawn are red is _________
Answer
0.37 to 0.38
Explanation
Let \(R_{1}\) and \(R_{2}\) be the events that the first and the second ball are red. Then
\[P(R_{1} \cap R_{2} \mid R_{1}) = \frac{P(R_{1} \cap R_{2})}{P(R_{1})} = \frac{\frac{10}{25} \times \frac{9}{24}}{\frac{10}{25}} = \frac{9}{24} = 0.375\]
In words: once a red ball is out, the bag holds 9 red and 15 blue balls, so the second ball is red with probability \(9/24\). The unconditional probability of two reds, \(\frac{10}{25} \times \frac{9}{24} = 0.15\), answers a different question.
Answer: 0.375