Difference between revisions of "1994 AHSME Problems/Problem 27"

(Solution)
 
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
Fix all your blank pages.. I'm sick of it. Get your stuff together AoPS...
+
==Problem==
 +
A bag of popping corn contains <math>\frac{2}{3}</math> white kernels and <math>\frac{1}{3}</math> yellow kernels. Only <math>\frac{1}{2}</math> of the white kernels will pop, whereas <math>\frac{2}{3}</math> 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?
 +
 
 +
<math> \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} </math>
 +
==Solution==
 +
To find the probability that the kernel is white, the probability of <math>P(\mathrm{white|popped}) = \frac{P(\mathrm{white, popped})}{P(\mathrm{popped})}</math>.
 +
 
 +
Running a bit of calculations <math>P(\mathrm{white, popped}) = \frac{1}{3}</math> while <math>P(\mathrm{popped}) = \frac{1}{3} + \frac{2}{9} = \frac{5}{9}</math>. Plugging this into the earlier equation, <math>P(\mathrm{white|popped}) = \frac{\frac{1}{3}}{\frac{5}{9}}</math>, meaning that the answer is <math>\boxed{\textbf{(D)}\ \frac{3}{5}}</math>.
 +
 
 +
==See Also==
 +
 
 +
{{AHSME box|year=1994|num-b=26|num-a=28}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:32, 9 January 2021

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png