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>. | ||
| − | 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)} [\frac{5}{12},\frac{2}{3}]}</math>. | + | 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>. |
==See Also== | ==See Also== | ||
Latest revision as of 12:52, 29 January 2023
Problem
The probability that event 
occurs is 
; the probability that event B occurs is 
.
Let 
be the probability that both and 
occur. The smallest interval necessarily containing is the interval
![$\textbf{(A)}\ \Big[\frac{1}{12},\frac{1}{2}\Big]\qquad \textbf{(B)}\ \Big[\frac{5}{12},\frac{1}{2}\Big]\qquad \textbf{(C)}\ \Big[\frac{1}{2},\frac{2}{3}\Big]\qquad \textbf{(D)}\ \Big[\frac{5}{12},\frac{2}{3}\Big]\qquad \textbf{(E)}\ \Big[\frac{1}{12},\frac{2}{3}\Big]$](http://latex.artofproblemsolving.com/a/7/3/a7397798ff92d0b1d2a3af9e2cd9382aaed49216.png)
Solution
Firstly note that 
and 
, as clearly the probability that both and occur cannot be more than the probability that or alone occurs. The more restrictive condition is , since 
.
Furthermore, by the Inclusion-Exclusion Principle, we also have ![\[\text{P}(A' \wedge B') = 1 - \text{P}(A) - \text{P}(B) + \text{P}(A \wedge B) = 1 - \frac{3}{4} - \frac{2}{3} + p = p - \frac{5}{12},\]](http://latex.artofproblemsolving.com/8/b/e/8be7b6a38825496d410b978913dc3f5d0bbb9474.png)
and as a probability must be non-negative, 
, so 
. Therefore, combining our inequalities gives 
, or ![$\boxed{\textbf{(D)} \Big[\frac{5}{12},\frac{2}{3}\Big]}$](http://latex.artofproblemsolving.com/c/0/2/c0220d6c3dbb87245fcdb7b09dea8b66de7af519.png)
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
