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.

Updated On: Jul 5, 2022
  • $\frac{32}{145}$
  • $\frac{28}{143}$
  • $\frac{36}{145}$
  • $\frac{36}{143}$
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The Correct Option is C

Solution and Explanation

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 $P(A\cap B) = P(A).P(B|A)=\frac{2}{5} \times \frac{18}{29}$ $= \frac{36}{145}$
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Uses of Semiconductor:

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