Question

A bag has 4 red and 6 black balls. Two balls are drawn without replacement. What is the probability that the second is red given the first was black?

• A. \( 4/10 \)
• B. \( 4/9 \)
• C. \( 6/9 \)
• D. \( 2/9 \)

Hint:

In conditional probability problems with “without replacement,” always update both the numerator and denominator after the first event occurs.

Answer 1

The Correct Option is B

Concept: Conditional probability is used when one event has already occurred. If sampling is done without replacement, the total number of objects decreases after each draw.

Step 1: Identify the initial composition of the bag:
Initially, \[ \text{Red balls} = 4, \qquad \text{Black balls} = 6 \] Total balls: \[ 4+6=10 \]

Step 2: Apply the given condition:
The first ball drawn is black. Hence one black ball is removed. Remaining balls: \[ \text{Red} = 4, \qquad \text{Black} = 5 \] Total remaining: \[ 4+5=9 \]

Step 3: Calculate the required probability:
Probability that the second ball is red: \[ P(\text{Second is Red} \mid \text{First is Black}) = \frac{4}{9} \]