Difference between revisions of "1996 AHSME Problems/Problem 16"
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==See also== | ==See also== | ||
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Revision as of 13:07, 5 July 2013
Problem
A fair standard six-sided dice is tossed three times. Given that the sum of the first two tosses equal the third, what is the probability that at least one "2" is tossed?
Solution
The third die cannot be 
, since the minimal sum on the other two dice is 
.
If the third die is , then the first two dice must be 
.
If the third die is 
, then the first two dice must be 
or 
.
If the third die is 
, then the first two dice must be 
, 
, or 
.
If the third die is 
, then the first two dice must be 
, 
, 
, or 
.
If the third die is 
, then the first two dice must be 
, 
, 
, 
, or 
.
There are 
possibilities for the three dice. Of those possibiltiies, 
have a in the first two dice, and has a in the third die. Therefore, the answer is 
.
See also
| 1996 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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. 
