2016 AMC 10B Problems/Problem 12
Problem
Two different numbers are selected at random from and multiplied together. What is the probability that the product is even?
Solution
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
.
Solution 2
There are 2 cases to get an even number. Case 1: even*even and Case 2: odd*even. Thus, to get an even*even, you get (2C2)/(5C2)= 1/10. And to get odd*even, you get [(3C1)*(2C1)]/(5C2) = 6/10. 1/10 + 6/10 yields 0.7 solution D
Video Solution
~savannahsolver
Video Solution
https://youtu.be/IRyWOZQMTV8?t=933
~ pi_is_3.14
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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