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.

1994 AMC 12 Problem 27

Answer: D

Difficulty rating: 1750

Small Hint:

Big Hint:

Solution:

Problem 27 in Other Years