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

Revision as of 22:16, 9 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

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

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

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

Revision as of 22:16, 9 January 2008

Problem

Solution

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Revision as of 11:31, 8 January 2008 (view source) 1=2 (talk \| contribs) ← Older edit	Revision as of 22:16, 9 January 2008 (view source) Azjps (talk \| contribs) m (1995 AMC 12 Problems/Problem 20 moved to 1995 AHSME Problems/Problem 20) Newer edit →
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