Question

A bag contains $12$ white pearls and $18$ black pearls. Two pearls are drawn in succession without replacement. Find the probability that the first pearl is white and the second is black.

• A. $\frac{32}{145}$
• B. $\frac{28}{143}$
• C. $\frac{36}{145}$
• D. $\frac{36}{143}$

Answer 1

Answer: (C)

$\frac{36}{145}$

Answer 2

Let $A$ and $B$ be the events of getting a white pearl in the first draw and a black pearl in the second draw. Now $P(A) = P$(getting a white pearl in the first draw) $=\frac{12}{30} = \frac{2}{5}$ When second pearl is drawn without replacement, the probability that the second pearl is black is the conditional probability of the event $B$ occurring when $A$ has already occurred. $\therefore P(B|A) = \frac{18}{29}$ By multiplication rule of probability, we have $P(A\cap B) = P(A).P(B|A)=\frac{2}{5} \times \frac{18}{29}$ $= \frac{36}{145}$