1994 AMC 12 Problem 27

Question

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?

\frac12

\frac59

\frac47

\frac35

\frac23

Answer: D

Concepts:conditional probability Bayes’ theorem

Difficulty rating: 1750

Small hint:

Compute the probabilities of selecting-and-popping for each color

Big hint:

Condition the white-and-popped probability on the total probability of popping

Solution:

The probabilities of selecting a kernel that pops are

\begin{gathered} P(W\text{ and pop})=\frac23\cdot\frac12=\frac13, \\ P(Y\text{ and pop})=\frac13\cdot\frac23=\frac29. \end{gathered}

Hence

P(\text{pop})=\frac{5}{9},

and

P(W\mid\text{pop})=\frac{\frac{1}{3}}{\frac{5}{9}}=\frac35.

Thus the correct answer is D.