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> | + | ==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 | ||
| + | \textbf{(B)}\ \frac13 \qquad | ||
| + | \textbf{(C)}\ \frac12 \qquad | ||
| + | \textbf{(D)}\ \frac23 \qquad | ||
| + | \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== | ||
| + | {{AHSME 30p box|year=1973|num-b=22|num-a=24}} | ||
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| + | [[Category:Introductory Probability Problems]] | ||
Latest revision as of 13:02, 20 February 2020
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
Solution
There are three red faces, and two are on the card that is completely red, so our answer is 
, which is 
.
See Also
| 1973 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
