A bag contains 4 red and 6 blue balls. Two balls are drawn one after the other without replacement. What is the probability that the second ball drawn is red, given that the first ball drawn was blue?
Show Hint
- \( \frac{4}{9} \)
- \( \frac{2}{5} \)
- \( \frac{1}{3} \)
- \( \frac{5}{9} \)
The Correct Option is A
Solution and Explanation
Step 1: Define the Problem
Total balls: 4 red + 6 blue = 10 balls. First ball drawn is blue (given). We need the probability that the second ball is red.
Step 2: Adjust for the First Draw
Since the first ball is blue, there are 6 blue balls initially, so the probability of drawing a blue ball first is \( \frac{6}{10} \). After drawing a blue ball, 9 balls remain: 4 red and 5 blue (since 6 - 1 = 5).
Step 3: Compute Conditional Probability
Probability that the second ball is red, given the first is blue, is the probability of drawing a red ball from the remaining 9 balls: \[ P(\text{second is red} \mid \text{first is blue}) = \frac{\text{Number of red balls}}{\text{Remaining balls}} = \frac{4}{9} \]
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