Difference between revisions of "1987 AJHSME Problems/Problem 25"
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<math>\text{(A)}\ \frac{4}{9} \qquad \text{(B)}\ \frac{9}{19} \qquad \text{(C)}\ \frac{1}{2} \qquad \text{(D)}\ \frac{10}{19} \qquad \text{(E)}\ \frac{5}{9}</math> | <math>\text{(A)}\ \frac{4}{9} \qquad \text{(B)}\ \frac{9}{19} \qquad \text{(C)}\ \frac{1}{2} \qquad \text{(D)}\ \frac{10}{19} \qquad \text{(E)}\ \frac{5}{9}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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. | 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. | ||
| Line 12: | Line 12: | ||
<math>\boxed{\text{A}}</math> | <math>\boxed{\text{A}}</math> | ||
| + | |||
| + | ==Solution 2 - Complementary Counting== | ||
| + | |||
| + | An alternative approach would be to calculate the probability the sum of the two numbers is odd, and subtract that from the total. We see that there is only one way to get an odd sum, where the two numbers must be of opposite parity. | ||
| + | |||
| + | Here, we have two cases <math>\rightarrow</math> the two numbers are, respectively, (even, odd), and (odd, even). The probability to get an even number as the first number is <math>\frac{5}{10} = \frac{1}{2},</math> and an odd number as the second number gives us a probability of <math>\frac{5}{9}.</math> Multiplying the two, we get <math>\frac{5}{18},</math> and because both cases have equal probability, we have the probability for the sum to be odd is <math>\frac{10}{18} = \frac{5}{9}.</math> | ||
| + | |||
| + | Finally, subtracting this from <math>1</math> gives us <math>\boxed{\frac{4}{9} \text{ } (A)}.</math> | ||
| + | ~ Cheetahboy93 | ||
==See Also== | ==See Also== | ||
{{AJHSME box|year=1987|num-b=24|after=Last<br>Problem}} | {{AJHSME box|year=1987|num-b=24|after=Last<br>Problem}} | ||
| − | [[Category:Probability Problems]] | + | [[Category:Introductory Probability Problems]] |
| + | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:05, 8 August 2024
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 1
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 
Solution 2 - Complementary Counting
An alternative approach would be to calculate the probability the sum of the two numbers is odd, and subtract that from the total. We see that there is only one way to get an odd sum, where the two numbers must be of opposite parity.
Here, we have two cases 
the two numbers are, respectively, (even, odd), and (odd, even). The probability to get an even number as the first number is 
and an odd number as the second number gives us a probability of 
Multiplying the two, we get 
and because both cases have equal probability, we have the probability for the sum to be odd is 
Finally, subtracting this from gives us 
~ Cheetahboy93
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
