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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>
  
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==Solution==
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==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.  
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<math>\boxed{\text{A}}</math>
 
<math>\boxed{\text{A}}</math>
  
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OR
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==Solution 2 - Complementary Counting==
  
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We find that it is only possible for the sum to be even if the numbers added are both even or odd. We will get an odd number when we add an even and an odd. We can use complementary counting to help solve the problem. There are a total of <math>90</math> possibilities since Jack can chose <math>10</math> numbers and Jill can pick <math>9</math>. The are <math>50</math> possibilities for the two numbers to be different since Jack can pick any of the <math>10</math> numbers and Jill has to pick from <math>5</math> numbers in the set with a different parity than the one that Jack picks. So the probability that the sum will be odd is <math>\frac{50}{90}=\frac{5}{9}</math>. Subtracting this by one gets the answer <math>\frac{4}{9}</math>
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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.
  
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<math>\boxed{\text{A}}</math>
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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>
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Finally, subtracting this from <math>1</math> gives us <math>\boxed{\frac{4}{9} \text{ } (A)}.</math>
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~ Cheetahboy93
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Probability Problems]]
 
[[Category:Introductory Probability Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 20:05, 8 August 2024

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 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 $\frac49$

$\boxed{\text{A}}$

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 $\rightarrow$ the two numbers are, respectively, (even, odd), and (odd, even). The probability to get an even number as the first number is $\frac{5}{10} = \frac{1}{2},$ and an odd number as the second number gives us a probability of $\frac{5}{9}.$ Multiplying the two, we get $\frac{5}{18},$ and because both cases have equal probability, we have the probability for the sum to be odd is $\frac{10}{18} = \frac{5}{9}.$

Finally, subtracting this from $1$ gives us $\boxed{\frac{4}{9} \text{ } (A)}.$ ~ Cheetahboy93

See Also

1987 AJHSME (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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