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Difference between revisions of "1983 AHSME Problems/Problem 26"

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Firstly note that <math>p \leq \frac{3}{4}</math> and <math>p \leq \frac{2}{3}</math>, as clearly the probability that both <math>A</math> and <math>B</math> occur cannot be more than the probability that <math>A</math> or <math>B</math> alone occurs. The more restrictive condition is <math>p \leq \frac{2}{3}</math>, since <math>\frac{2}{3} < \frac{3}{4}</math>.  
 
Firstly note that <math>p \leq \frac{3}{4}</math> and <math>p \leq \frac{2}{3}</math>, as clearly the probability that both <math>A</math> and <math>B</math> occur cannot be more than the probability that <math>A</math> or <math>B</math> alone occurs. The more restrictive condition is <math>p \leq \frac{2}{3}</math>, since <math>\frac{2}{3} < \frac{3}{4}</math>.  
  
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Now, by the Inclusion-Exclusion Principle, we also have <cmath>\text{P}(A' \wedge B') = 1 - \text{P}(A) - \text{P}(B) + \text{P}(A \wedge B) = 1 - \frac{3}{4} - \frac{2}{3} + p = p - \frac{5}{12},</cmath> and as a probability must be non-negative, <math>p - \frac{5}{12} \geq 0</math>, so <math>p \geq \frac{5}{12}</math>. Therefore, combining our inequalities gives <math>\frac{5}{12} \leq p \leq \frac{2}{3}</math>, which is answer choice <math>\boxed{\textbf{D}}</math>.
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Furthermore, by the [[Inclusion-Exclusion Principle]], we also have <cmath>\text{P}(A' \wedge B') = 1 - \text{P}(A) - \text{P}(B) + \text{P}(A \wedge B) = 1 - \frac{3}{4} - \frac{2}{3} + p = p - \frac{5}{12},</cmath> and as a probability must be non-negative, <math>p - \frac{5}{12} \geq 0</math>, so <math>p \geq \frac{5}{12}</math>. Therefore, combining our inequalities gives <math>\frac{5}{12} \leq p \leq \frac{2}{3}</math>, or <math>\boxed{\textbf{(D)} \Big[\frac{5}{12},\frac{2}{3}\Big]}</math>.
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==See Also==
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{{AHSME box|year=1983|num-b=25|num-a=27}}
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{{MAA Notice}}

Latest revision as of 12:52, 29 January 2023

Problem

The probability that event $A$ occurs is $\frac{3}{4}$; the probability that event B occurs is $\frac{2}{3}$. Let $p$ be the probability that both $A$ and $B$ occur. The smallest interval necessarily containing $p$ is the interval

$\textbf{(A)}\ \Big[\frac{1}{12},\frac{1}{2}\Big]\qquad \textbf{(B)}\ \Big[\frac{5}{12},\frac{1}{2}\Big]\qquad \textbf{(C)}\ \Big[\frac{1}{2},\frac{2}{3}\Big]\qquad \textbf{(D)}\ \Big[\frac{5}{12},\frac{2}{3}\Big]\qquad \textbf{(E)}\ \Big[\frac{1}{12},\frac{2}{3}\Big]$

Solution

Firstly note that $p \leq \frac{3}{4}$ and $p \leq \frac{2}{3}$, as clearly the probability that both $A$ and $B$ occur cannot be more than the probability that $A$ or $B$ alone occurs. The more restrictive condition is $p \leq \frac{2}{3}$, since $\frac{2}{3} < \frac{3}{4}$.

Furthermore, by the Inclusion-Exclusion Principle, we also have \[\text{P}(A' \wedge B') = 1 - \text{P}(A) - \text{P}(B) + \text{P}(A \wedge B) = 1 - \frac{3}{4} - \frac{2}{3} + p = p - \frac{5}{12},\] and as a probability must be non-negative, $p - \frac{5}{12} \geq 0$, so $p \geq \frac{5}{12}$. Therefore, combining our inequalities gives $\frac{5}{12} \leq p \leq \frac{2}{3}$, or $\boxed{\textbf{(D)} \Big[\frac{5}{12},\frac{2}{3}\Big]}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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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