Difference between revisions of "2015 AMC 8 Problems/Problem 7"
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+ | ==Problem== | ||
+ | |||
Each of two boxes contains three chips numbered <math>1</math>, <math>2</math>, <math>3</math>. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even? | Each of two boxes contains three chips numbered <math>1</math>, <math>2</math>, <math>3</math>. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even? | ||
<math>\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}</math> | <math>\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}</math> | ||
− | ===Solution=== | + | ==Solutions== |
+ | ===Solution 1=== | ||
We can instead calculate the probability that their product is odd, and subtract this from <math>1</math>. In order to get an odd product, we have to draw an odd number from each box. We have a <math>\frac{2}{3}</math> probability of drawing an odd number from one box, so there is a <math>\left ( \frac{2}{3} \right )^2=\frac{4}{9}</math> probability of having an odd product. Thus, there is a <math>1-\frac{4}{9}=\boxed{\textbf{(E)}~\frac{5}{9}}</math> probability of having an even product. | We can instead calculate the probability that their product is odd, and subtract this from <math>1</math>. In order to get an odd product, we have to draw an odd number from each box. We have a <math>\frac{2}{3}</math> probability of drawing an odd number from one box, so there is a <math>\left ( \frac{2}{3} \right )^2=\frac{4}{9}</math> probability of having an odd product. Thus, there is a <math>1-\frac{4}{9}=\boxed{\textbf{(E)}~\frac{5}{9}}</math> probability of having an even product. | ||
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<math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | We can also list out the numbers. Hat A has chips <math>1</math>, <math>2</math>, and <math>3</math>, and Hat B also has chips <math>1</math>, <math>2</math>, and <math>3</math>. Chip <math>1</math>(from Hat A) | ||
+ | |||
+ | could be with 3 partners from Hat B. This is also the same for chips <math>2</math> and <math>3</math> from Hat A. <math>3+3+3=9</math> total sums. Chip <math>1</math> could be | ||
+ | multiplied with 2 other chips to make an even product, just like chip <math>3</math>. Chip <math>2</math> can only multiply with 1 chip. <math>2+2+1=5</math>. The answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | ||
+ | |||
+ | ===Solution 4=== | ||
+ | Here is another way: | ||
+ | |||
+ | Let's start by finding the denominator: Total choices. | ||
+ | There are <math>3</math> chips we can choose from in the 1st box, and <math>3</math> chips we can choose from in the 2nd box. We do <math>3*3</math>, and get <math>9</math>. | ||
+ | Now - to find the numerator: Desired choices. | ||
+ | To get an even number, we need to pick 2 from at least one of the boxes. There are <math>2</math> choices as to finding which box we will draw the 2 from. Then we have <math>3</math> choices from the other box to pick any of the other chips, <math>1, 2, 3</math>. | ||
+ | |||
+ | <math>\frac{3 \cdot2}{9} = \frac{6}{9}</math> | ||
+ | However, we are over counting, the <math>(2,2)</math> configuration twice, and so we subtract that one configuration from our total. | ||
+ | <math>\frac{6}{9} - \frac{1}{9}</math>. | ||
+ | Thus, our answer is <math>\boxed{\textbf{(E) }\frac{5}{9}}</math>. | ||
+ | |||
+ | ~ del-math. | ||
==See Also== | ==See Also== |
Latest revision as of 16:35, 16 January 2021
Contents
Problem
Each of two boxes contains three chips numbered , , . A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?
Solutions
Solution 1
We can instead calculate the probability that their product is odd, and subtract this from . In order to get an odd product, we have to draw an odd number from each box. We have a probability of drawing an odd number from one box, so there is a probability of having an odd product. Thus, there is a probability of having an even product.
Solution 2
You can also make this problem into a spinner problem. You have the first spinner with equally divided
sections, and You make a second spinner that is identical to the first, with equal sections of
,, and . If the first spinner lands on , to be even, it must land on two. You write down the first
combination of numbers . Next, if the spinner lands on , it can land on any number on the second
spinner. We now have the combinations of . Finally, if the first spinner ends on , we
have Since there are possible combinations, and we have evens, the final answer is
.
Solution 3
We can also list out the numbers. Hat A has chips , , and , and Hat B also has chips , , and . Chip (from Hat A)
could be with 3 partners from Hat B. This is also the same for chips and from Hat A. total sums. Chip could be multiplied with 2 other chips to make an even product, just like chip . Chip can only multiply with 1 chip. . The answer is .
Solution 4
Here is another way:
Let's start by finding the denominator: Total choices. There are chips we can choose from in the 1st box, and chips we can choose from in the 2nd box. We do , and get . Now - to find the numerator: Desired choices. To get an even number, we need to pick 2 from at least one of the boxes. There are choices as to finding which box we will draw the 2 from. Then we have choices from the other box to pick any of the other chips, .
However, we are over counting, the configuration twice, and so we subtract that one configuration from our total. . Thus, our answer is .
~ del-math.
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.