Question

One bag contains 4 white balls and 2 black balls. Another bag contains 3 white balls and 5 black balls. If one ball is drawn from each bag, determine the probability that one ball is white and another is black.

• A. $\frac{6}{24}$
• B. $\frac{5}{24}$
• C. $\frac{7}{24}$
• D. $\frac{13}{24}$

Answer 1

The correct option is D $\frac{13}{24}$

Probability that first ball is white and second black = $\left(4/6\right)\times \left(5/8\right)=5/12$
Probability that first ball is black and second white =$\left(2/6\right)\times \left(3/8\right)=1/8$
These are mutually exclusive events hence the required probability
$P=\frac{5}{12}+\frac{1}{8}=\frac{13}{24}$.