2015 AMC 8 Problems/Problem 7
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 , it must land on two for the result to be even. 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 and . 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. Box A has chips , , and , and Box B also has chips , , and . Chip (from Box A)
could be with 3 partners from Box B. This is also the same for chips and from Box 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, and .
However, we are over counting the configuration twice, and so, we subtract that one configuration from our total. . Thus, our answer is .
~ del-math.
Solution 5
This might take longer to solve, but you definitely will have the right answer. You first list out all the possible combinations regardless if the product is even or odd. There are 9 possible combinations, and of the combinations’ products are even. So, our answer is .
~MiracleMaths
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
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.