Difference between revisions of "1995 AHSME Problems/Problem 20"
(@azjps: THANK YOU THANK YOU THANK YOU for putting solution + image up!) |
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If <math>a,b</math> and <math>c</math> are three (not necessarily different) numbers chosen randomly and with replacement from the set <math>\{1,2,3,4,5 \}</math>, the probability that <math>ab + c</math> is even is | If <math>a,b</math> and <math>c</math> are three (not necessarily different) numbers chosen randomly and with replacement from the set <math>\{1,2,3,4,5 \}</math>, the probability that <math>ab + c</math> is even is | ||
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| − | <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> | + | <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 == | == Solution == | ||
Revision as of 11:31, 8 January 2008
Problem
If 
and 
are three (not necessarily different) numbers chosen randomly and with replacement from the set 
, the probability that 
is even is
Solution
The probability of 
being odd is 
, so the probability of being even is 
.
The probability of being odd is 
and being even is 
.
is even if and are both odd, with probability 
; or and are both even, with probability 
. Thus the total probability is 
.
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 | ||
