1983 AMC 12 Problem 26

Attempt Problem 26 of the 1983 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1983 AMC 12 solutions, or check the answer key.

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26.

The probability that event AA occurs is 34;\frac34; the probability that event BB occurs is 23.\frac23. Let pp be the probability that both AA and BB occur. The smallest interval necessarily containing pp is the interval
[112,12][\frac1{12},\frac12]
[512,12][\frac5{12},\frac12]
[12,23][\frac12,\frac23]
[512,23][\frac5{12},\frac23]
[112,23][\frac1{12},\frac23]
Answer: D
Concepts:inclusion-exclusioninequality
Small hint:
Apply inclusion-exclusion to events AA and BB
Big hint:
Bound P(A∪B)P(A\cup B) between max⁡(P(A),P(B))\max(P(A),P(B)) and 11
Solution:
Inclusion-exclusion gives p=P(A∩B)=34+23−P(A∪B). \begin{aligned} p&=P(A\cap B)\\ &=\frac34+\frac23-P(A\cup B). \end{aligned} Since 34≤P(A∪B)≤1,\frac{3}{4}\le P(A\cup B)\le1, it follows that 512≤p≤23. \frac5{12}\le p\le\frac23. Both endpoints can occur, so this is the smallest necessary interval.
Therefore, the correct answer is D.
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