Difference between revisions of "1987 AJHSME Problems/Problem 25"

(New page: ==Problem== Ten balls numbered <math>1</math> to <math>10</math> are in a jar. Jack reaches into the jar and randomly removes one of the balls. Then Jill reaches into the jar and random...)
 
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==Problem==
 
==Problem==
  
Ten balls numbered <math>1</math> to <math>10</math> 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  
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Ten balls numbered <math>1</math> to <math>10</math> 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  
  
 
<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>
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==Solution==
 
==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.  
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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 <math>\frac49</math>
 
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 <math>\frac49</math>
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==See Also==
 
==See Also==
  
[[1987 AJHSME Problems]]
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{{AJHSME box|year=1987|num-b=24|after=Last<br>Problem}}
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[[Category:Probability Problems]]

Revision as of 08:04, 31 May 2009

Problem

Ten balls numbered $1$ to $10$ 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

$\text{(A)}\ \frac{4}{9} \qquad \text{(B)}\ \frac{9}{19} \qquad \text{(C)}\ \frac{1}{2} \qquad \text{(D)}\ \frac{10}{19} \qquad \text{(E)}\ \frac{5}{9}$

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 $\frac49$

$\boxed{\text{A}}$

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
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