Question

A drawer contains black socks and white socks. A person randomly selects two socks without replacement. The probability of selecting two black socks is 120703 and the probability of selecting a black sock on the first draw is 819. What is the probability of selecting a black sock on the second draw given that a black sock was selected on the first draw

Answer 1

Answer:

0.4054 = 40.54% probability of selecting a black sock on the second draw given that a black sock was selected on the first draw

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Black sock on the first draw.

Event B: Black sock on the second draw.

The probability of selecting a black sock on the first draw is 8/19.

This means that $P(A) = \frac{8}{19}$

Black socks on both draws:

The probability of selecting two black socks is 120/703

This means that $P(A \cap B) = \frac{120}{703}$

What is the probability of selecting a black sock on the second draw given that a black sock was selected on the first draw?

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

$P(B|A) = \frac{\frac{120}{703}}{\frac{8}{19}}$

$P(B|A) = \frac{120}{703}*\frac{19}{8}$

$P(B|A) = \frac{120*19}{703*8}$

$P(B|A) = 0.4054$

0.4054 = 40.54% probability of selecting a black sock on the second draw given that a black sock was selected on the first draw