Question

A bag contains \( 5 \) red and \( 4 \) black balls. Two balls are drawn at random one after the other without replacement. What is the conditional probability that the second ball drawn is red, given that the first ball drawn was black?

• A. \( \frac{5}{8} \)
• B. \( \frac{5}{9} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{4}{9} \)

Hint:

For simple dependent sequence probability questions, adjust the numerator and denominator parameters based on the condition given instead of working out lengthy Bayes' formula expansions.

Answer 1

The Correct Option is A

Concept: Conditional probability deals with finding the likelihood of an event occurring given that a prerequisite event has already taken place. This alters the sample space configuration for the subsequent event.

Step 1: Analyze the initial composition of the bag. Initially, the bag contains:
• Red balls = \( 5 \)
• Black balls = \( 4 \)
• Total number of balls = \( 5 + 4 = 9 \)

Step 2: Update the bag contents after the first event occurs. The condition states that the first ball drawn was black. Since the draw is performed without replacement, we remove one black ball from the bag:
• Remaining Red balls = \( 5 \) (unchanged)
• Remaining Black balls = \( 4 - 1 = 3 \)
• New Total number of balls = \( 5 + 3 = 8 \)

Step 3: Calculate the probability of drawing a red ball from the updated sample space. The probability that the second ball is red given the first was black is: \[ P(\text{Red}_2 | \text{Black}_1) = \frac{\text{Number of remaining red balls}}{\text{New total number of balls}} = \frac{5}{8} \]