Independent solution

How to solve this Conditional Probability question

Setup

Setup

Let E be the event that the transferred sample contains two balls of each color and let B be the event that the subsequent selection is blue.

Pr⁡(E)=(82)(62)(144)\Pr(E)=\frac{\binom82\binom62}{\binom{14}4}

Model

Model

Under E, half of the transferred balls are blue. The marginal selected color matches a uniformly selected original ball.

Pr⁡(B∣E)=24=12\Pr(B\mid E)=\frac24=\frac12
Pr⁡(B)=614=37\Pr(B)=\frac6{14}=\frac37

Compute

Compute

Apply Bayes' rule using the joint numerator.

Pr⁡(E∩B)=(82)(62)(144)(12)=0.2097902\Pr(E\cap B)=\frac{\binom82\binom62}{\binom{14}4}\left(\frac12\right)=0.2097902
Pr⁡(E∣B)=0.20979023/7=0.4895105\Pr(E\mid B)=\frac{0.2097902}{3/7}=0.4895105

Answer

Answer

The conditional probability rounds to 0.49.

0.49(D)\boxed{0.49\quad\text{(D)}}