Difference between revisions of "1995 AHSME Problems/Problem 20"

Revision as of 10:30, 8 January 2008

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

A. $\frac {2}{5}$ B. $\frac {59}{125}$ C. $\frac {1}{2}$ D. $\frac {64}{125}$ E. $\frac {3}{5}$

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

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 $\frac{59}{125} \Rightarrow \mathrm{(B)}$ .

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

@@ Line 8: / Line 8: @@
 D. <math>\frac {64}{125}</math>
 E. <math>\frac {3}{5}</math>
+<math> \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} }  </math>
 == Solution ==

Page

Toolbox

Search

Difference between revisions of "1995 AHSME Problems/Problem 20"

Revision as of 10:30, 8 January 2008

Problem

Solution

See also