Question:
A box contains \(5\) red and \(4\) black balls. Two balls are drawn one by one without replacement. What is the probability that the second ball is red, given that the first ball drawn was black?
A box contains \(5\) red and \(4\) black balls. Two balls are drawn one by one without replacement. What is the probability that the second ball is red, given that the first ball drawn was black?
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In "without replacement" problems, mentally update the inventory of the urn/box immediately after the given condition is stated before computing the final probability.
Updated On: Jun 15, 2026
- \(5/8\)
- \(4/9\)
- \(5/9\)
- \(1/2\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
This is a problem based on conditional probability. We need to find the probability of a specific outcome in the second draw (getting a red ball), given the certainty of the outcome of the first draw (getting a black ball) without replacing it.
Step 2: Key Formula or Approach:
Because the draws are without replacement, the outcome of the first draw affects the total number of balls and the composition of the box for the second draw.
The conditional probability of an event \(A\) given that event \(B\) has already occurred is evaluated by counting the remaining favorable outcomes divided by the new total number of outcomes.
Step 3: Detailed Explanation:
Initially, the composition of the box is:
Number of red balls = 5
Number of black balls = 4
Total number of balls = 9
We are given that the first event has already occurred:
The first ball drawn is black.
Since the ball is drawn without replacement, we must remove one black ball from the initial counts.
The new composition of the box before the second draw is:
Number of red balls remaining = 5 (none were removed)
Number of black balls remaining = \(4 - 1 = 3\)
Total number of balls remaining = \(5 + 3 = 8\)
Now, we want to find the probability of drawing a
red ball from this newly composed box.
\[ P(\text{Second is Red} \mid \text{First is Black}) = \frac{\text{Number of Red balls remaining}}{\text{Total number of balls remaining}} \] \[ P(\text{Second is Red} \mid \text{First is Black}) = \frac{5}{8} \] (Note: The options in the source image are cut off. Plausible options have been reconstructed for structural completeness.)
Step 4: Final Answer:
The correct choice is (A).
This is a problem based on conditional probability. We need to find the probability of a specific outcome in the second draw (getting a red ball), given the certainty of the outcome of the first draw (getting a black ball) without replacing it.
Step 2: Key Formula or Approach:
Because the draws are without replacement, the outcome of the first draw affects the total number of balls and the composition of the box for the second draw.
The conditional probability of an event \(A\) given that event \(B\) has already occurred is evaluated by counting the remaining favorable outcomes divided by the new total number of outcomes.
Step 3: Detailed Explanation:
Initially, the composition of the box is:
Number of red balls = 5
Number of black balls = 4
Total number of balls = 9
We are given that the first event has already occurred:
The first ball drawn is black.
Since the ball is drawn without replacement, we must remove one black ball from the initial counts.
The new composition of the box before the second draw is:
Number of red balls remaining = 5 (none were removed)
Number of black balls remaining = \(4 - 1 = 3\)
Total number of balls remaining = \(5 + 3 = 8\)
Now, we want to find the probability of drawing a
red ball from this newly composed box.
\[ P(\text{Second is Red} \mid \text{First is Black}) = \frac{\text{Number of Red balls remaining}}{\text{Total number of balls remaining}} \] \[ P(\text{Second is Red} \mid \text{First is Black}) = \frac{5}{8} \] (Note: The options in the source image are cut off. Plausible options have been reconstructed for structural completeness.)
Step 4: Final Answer:
The correct choice is (A).
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