Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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There are two different cases in which the product is zero; either the first number we select is zero, or the second one is. We consider these cases separately. | There are two different cases in which the product is zero; either the first number we select is zero, or the second one is. We consider these cases separately. | ||
− | Case 1: 0 is the first number chosen | + | Case 1: <math>0</math> is the first number chosen |
There is a <math>\frac{1}{6}</math> chance of selecting zero as the first number. At this point, the product will be zero no matter the choice of the second number, so there is a <math>\frac{1}{6}</math> of getting the desired product in this case. | There is a <math>\frac{1}{6}</math> chance of selecting zero as the first number. At this point, the product will be zero no matter the choice of the second number, so there is a <math>\frac{1}{6}</math> of getting the desired product in this case. | ||
− | Case 2: 0 is the second number chosen | + | Case 2: <math>0</math> is the second number chosen |
There is a <math>\frac{5}{6}</math> chance of choosing a number that is NOT zero as the first number. From there, there is a <math>\frac{1}{5}</math> chance of picking zero from the remaining 5 numbers. Thus, there is a <math>\frac{5}{6} \cdot \frac{1}{5} = \frac{1}{6}</math> chance of getting a product of 0 in this case. | There is a <math>\frac{5}{6}</math> chance of choosing a number that is NOT zero as the first number. From there, there is a <math>\frac{1}{5}</math> chance of picking zero from the remaining 5 numbers. Thus, there is a <math>\frac{5}{6} \cdot \frac{1}{5} = \frac{1}{6}</math> chance of getting a product of 0 in this case. |
Revision as of 01:18, 15 January 2024
Contents
Problem
Two different numbers are randomly selected from the set and multiplied together. What is the probability that the product is
?
Solutions
Solution 1
The product can only be if one of the numbers is
. Once we chose
, there are
ways we can chose the second number, or
. There are
ways we can chose
numbers randomly, and that is
. So,
so the answer is
.
Solution 2 (Complementary Counting)
Because the only way the product of the two numbers is is if one of the numbers we choose is
we calculate the probability of NOT choosing a
We get
Therefore our answer is
Solution 3 (Casework)
There are two different cases in which the product is zero; either the first number we select is zero, or the second one is. We consider these cases separately.
Case 1: is the first number chosen
There is a chance of selecting zero as the first number. At this point, the product will be zero no matter the choice of the second number, so there is a
of getting the desired product in this case.
Case 2: is the second number chosen
There is a chance of choosing a number that is NOT zero as the first number. From there, there is a
chance of picking zero from the remaining 5 numbers. Thus, there is a
chance of getting a product of 0 in this case.
Adding the probabilities from the two distinct cases up, we find that there is a chance of getting a product of zero.
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=357
~ pi_is_3.14
Video Solution
~savannahsolver
See Also
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.