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. \(5/8\)
• B. \(4/9\)
• C. \(5/9\)
• D. \(1/2\)

Hint:

In "without replacement" problems, mentally update the inventory of the urn/box immediately after the given condition is stated before computing the final probability.

Answer 1

The Correct Option is A

Step 1: Understand the problem.
A box holds $5$ red and $4$ black balls. Two balls are drawn one after another without replacement. We want the chance the second ball is red, given that the first was black.
Step 2: Note the starting counts.
Red $= 5$, black $= 4$, total $= 9$ balls.
Step 3: Use the given condition.
We are told the first ball drawn was black, and it is not put back. So one black ball leaves the box for good.
Step 4: Update the box.
Reds remain $5$, blacks drop to $4 - 1 = 3$, and the total becomes $5 + 3 = 8$ balls.
Step 5: Set up the conditional probability.
Now drawing from the updated box, \[ P(\text{2nd red} \mid \text{1st black}) = \frac{\text{red balls left}}{\text{total balls left}} = \frac{5}{8} \]
Step 6: State the answer.
The required probability is $\dfrac{5}{8}$, which is option (A).
\[ \boxed{P = \dfrac{5}{8}} \]