1994 AHSME Problems/Problem 27

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Problem

A bag of popping corn contains $\frac{2}{3}$ white kernels and $\frac{1}{3}$ yellow kernels. Only $\frac{1}{2}$ of the white kernels will pop, whereas $\frac{2}{3}$ of the yellow ones will pop. A kernel is selected at random from the bag, and pops when placed in the popper. What is the probability that the kernel selected was white?

$\textbf{(A)}\ \frac{1}{2} \qquad\textbf{(B)}\ \frac{5}{9} \qquad\textbf{(C)}\ \frac{4}{7} \qquad\textbf{(D)}\ \frac{3}{5} \qquad\textbf{(E)}\ \frac{2}{3}$

Solution

To find the probability that the kernel is white, the probability of $P(\mathrm{white|popped}) = \frac{P(\mathrm{white, popped})}{P(\mathrm{popped})}$.

Running a bit of calculations $P(\mathrm{white, popped}) = \frac{1}{3}$ while $P(\mathrm{popped}) = \frac{1}{3} + \frac{2}{9} = \frac{5}{9}$. Plugging this into the earlier equation, $P(\mathrm{white|popped}) = \frac{\frac{1}{3}}{\frac{5}{9}}$, meaning that the answer is $\boxed{\textbf{(D)}\ \frac{3}{5}}$.

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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