2016 AMC 10B Problems/Problem 12
Contents
Problem
Two different numbers are selected at random from and multiplied together. What is the probability that the product is even?
Solution 1
The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is , so the answer is which is .
An alternate way to finish: Since it is odd if none are even, the probability is . ~Alternate solve by JH. L
Solution 2
There are cases to get an even number. Case 1: and Case 2: . Thus, to get an , you get . And to get , you get . which is and the answer is .
Solution 3
Note that we have three cases to get an even number: even even, odd even and even odd. The probability of case 1 is , the probability of case 2 is and the probability of case 3 is .
Adding these up we get
-ConfidentKoala4
Video Solution by OmegaLearn
https://youtu.be/IRyWOZQMTV8?t=933
~ pi_is_3.14
Video Solution
https://youtu.be/tUpKpGmOwDQ - savannahsolver
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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All AMC 10 Problems and Solutions |
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