2015 AMC 8 Problems/Problem 7
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?
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 .
Solution 4 by: 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 |
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