Difference between revisions of "1994 AHSME Problems/Problem 27"
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− | + | ==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? | ||
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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> | ||
+ | ==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>. | ||
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+ | 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}} | {{MAA Notice}} |
Latest revision as of 16:32, 9 January 2021
Problem
A bag of popping corn contains white kernels and yellow kernels. Only of the white kernels will pop, whereas 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?
Solution
To find the probability that the kernel is white, the probability of .
Running a bit of calculations while . Plugging this into the earlier equation, , meaning that the answer is .
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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All AHSME Problems and Solutions |
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