Difference between revisions of "1973 AHSME Problems/Problem 23"

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There are three red faces, and two are on the card that is completely red, so our answer is <math>\frac{2}{3}</math>, which is <math>\boxed{D}</math>.
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==Problem==
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There are two cards; one is red on both sides and the other is red on one side and blue on the other. The cards have the same probability (1/2) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is red, then the probability that the under-side is also red is
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<math> \textbf{(A)}\ \frac14 \qquad
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\textbf{(B)}\ \frac13 \qquad
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\textbf{(C)}\ \frac12 \qquad
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\textbf{(D)}\ \frac23 \qquad
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\textbf{(E)}\ \frac34</math>
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==Solution==
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There are three red faces, and two are on the card that is completely red, so our answer is <math>\frac{2}{3}</math>, which is <math>\boxed{\textbf{(D)}}</math>.
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==See Also==
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{{AHSME 35p box|year=1973|num-b=22|num-a=24}}
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[[Category:Introductory Probability Problems]]

Revision as of 11:20, 5 July 2018

Problem

There are two cards; one is red on both sides and the other is red on one side and blue on the other. The cards have the same probability (1/2) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is red, then the probability that the under-side is also red is

$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac12 \qquad \textbf{(D)}\ \frac23 \qquad \textbf{(E)}\ \frac34$

Solution

There are three red faces, and two are on the card that is completely red, so our answer is $\frac{2}{3}$, which is $\boxed{\textbf{(D)}}$.

See Also

1973 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AHSME Problems and Solutions