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Difference between revisions of "1995 AHSME Problems/Problem 20"
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The probability of <math>ab</math> being odd is <math>\left(\frac 35\right)^2 = \frac{9}{25}</math>, so the probability of <math>ab</math> being even is <math>1 - \frac{9}{25} = \frac {16}{25}</math>.
 
The probability of <math>ab</math> being odd is <math>\left(\frac 35\right)^2 = \frac{9}{25}</math>, so the probability of <math>ab</math> being even is <math>1 - \frac{9}{25} = \frac {16}{25}</math>.
  
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The probability of <math>c</math> being odd is <math>3/5</math> and being even is <math>2/5</math>.
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The probability of <math>c</math> being odd is <math>3/5</math> and being even is <math>2/5</math>
  
 
<math>ab+c</math> is even if <math>ab</math> and <math>c</math> are both odd, with probability <math>\frac{9}{25} \cdot \frac{3}{5} = \frac{27}{125}</math>; or <math>ab</math> and <math>c</math> are both even, with probability <math>\frac{16}{25} \cdot \frac{2}{5} = \frac{32}{125}</math>. Thus the total probability is <math>\frac{59}{125} \Rightarrow \mathrm{(B)}</math>.
 
<math>ab+c</math> is even if <math>ab</math> and <math>c</math> are both odd, with probability <math>\frac{9}{25} \cdot \frac{3}{5} = \frac{27}{125}</math>; or <math>ab</math> and <math>c</math> are both even, with probability <math>\frac{16}{25} \cdot \frac{2}{5} = \frac{32}{125}</math>. Thus the total probability is <math>\frac{59}{125} \Rightarrow \mathrm{(B)}</math>.
  
 
== See also ==
 
== See also ==
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{{Old AMC12 box|year=1995|num-b=19|num-a=21}}
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{{AHSME box|year=1995|num-b=19|num-a=21}}
  
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[[Category:Introductory Probability Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 17:05, 29 July 2022

Problem

If $a,b$ and $c$ are three (not necessarily different) numbers chosen randomly and with replacement from the set $\{1,2,3,4,5 \}$, the probability that $ab + c$ is even is


$\mathrm{(A) \ \frac {2}{5} } \qquad \mathrm{(B) \ \frac {59}{125} } \qquad \mathrm{(C) \ \frac {1}{2} } \qquad \mathrm{(D) \ \frac {64}{125} } \qquad \mathrm{(E) \ \frac {3}{5} }$

Solution

The probability of $ab$ being odd is $\left(\frac 35\right)^2 = \frac{9}{25}$, so the probability of $ab$ being even is $1 - \frac{9}{25} = \frac {16}{25}$.

The probability of $c$ being odd is $3/5$ and being even is $2/5$

$ab+c$ is even if $ab$ and $c$ are both odd, with probability $\frac{9}{25} \cdot \frac{3}{5} = \frac{27}{125}$; or $ab$ and $c$ are both even, with probability $\frac{16}{25} \cdot \frac{2}{5} = \frac{32}{125}$. Thus the total probability is $\frac{59}{125} \Rightarrow \mathrm{(B)}$.

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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