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

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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>
+
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>. There 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>
  
 
<math>\boxed{\text{A}}</math>
 
<math>\boxed{\text{A}}</math>

Revision as of 16:11, 2 August 2012

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 $\frac49$

$\boxed{\text{A}}$

OR

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 $90$ possibilities since Jack can chose $10$ numbers and Jill can pick $9$. There are $50$ possibilities for the two numbers to be different since Jack can pick any of the $10$ numbers and Jill has to pick from $5$ numbers in the set with a different parity than the one that Jack picks. So the probability that the sum will be odd is $\frac{50}{90}=\frac{5}{9}$. Subtracting this by one gets the answer $\frac{4}{9}$

$\boxed{\text{A}}$

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
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