Difference between revisions of "1987 AJHSME Problems/Problem 25"
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{{AJHSME box|year=1987|num-b=24|after=Last<br>Problem}} | {{AJHSME box|year=1987|num-b=24|after=Last<br>Problem}} | ||
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| + | [[Category:Introductory Combinatorics Problems]] | ||
Revision as of 05:52, 16 April 2012
Problem
Ten balls numbered 
to 
are in a jar. Jack reaches into the jar and randomly removes one of the balls. Then Jill reaches into the jar and randomly removes a different ball. The probability that the sum of the two numbers on the balls removed is even is
Solution
For the sum of the two numbers removed to be even, they must be of the same parity. There are five even values and five odd values.
No matter what Jack chooses, the number of numbers with the same parity is four. There are nine numbers total, so the probability Jill chooses a number with the same parity as Jack's is 
See Also
| 1987 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
