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

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<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>
 
<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==
 
==Solution==
−
To find the probability that the kernel is white, the probability of <math>P(white|popped) = \frac{P(white, popped)}{P(popped)}</math>
+
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(white, popped) = \frac{1}{3}</math> while <math>P(popped) = \frac{1}{3} + \frac{2}{9} = \frac{5}{9}</math> Plugging this into the earlier equation, <math>P(white|popped) = \frac{\frac{1}{3}}{\frac{5}{9}}</math>. Meaning that the answer is <math>\boxed{\textbf{(D)}\ \frac{3}{5}}</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>.
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==See Also==
 +
 
 +
{{AHSME box|year=1994|num-b=26|num-a=28}}
 +
{{MAA Notice}}

Latest revision as of 16:32, 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 (Problems • Answer Key • Resources)
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

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