Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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− | Two different numbers are randomly selected from the set <math>{ - 2, -1, 0, 3, 4, 5}</math> and multiplied together. What is the probability that the product is <math>0</math>? | + | == Problem == |
+ | |||
+ | Two different numbers are randomly selected from the set <math>\{ - 2, -1, 0, 3, 4, 5\}</math> and multiplied together. What is the probability that the product is <math>0</math>? | ||
<math>\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}</math> | <math>\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}</math> | ||
− | ==Solution== | + | == Solutions== |
− | The | + | |
+ | ===Solution 1=== | ||
+ | 1. Identify the total number of ways to select two different numbers from the set: | ||
+ | |||
+ | The set has 6 elements. The number of ways to choose 2 different numbers from 6 is given by the combination formula: <math>\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15</math>. | ||
+ | |||
+ | 2. Identify the favorable outcomes: | ||
+ | |||
+ | For the product to be zero, one of the chosen numbers must be zero. The set contains one zero (0). To have a product of zero, we need to choose 0 and any other number from the remaining five numbers <math>-2, -1, 3, 4, 5</math>. | ||
+ | |||
+ | The number of ways to choose 0 and one other number from the remaining five is 5. | ||
+ | |||
+ | 3. Calculate the probability: | ||
+ | |||
+ | The probability is the number of favorable outcomes divided by the total number of outcomes: <math>\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{15} = \frac{1}{3}</math>. | ||
+ | |||
+ | Thus, the probability that the product is <math>0</math> is <math>\boxed{\textbf{(D)} \ \frac{1}{3}}.</math> | ||
+ | |||
+ | ~GeometryMystery | ||
+ | |||
+ | ===Solution 2 (Complementary Counting)=== | ||
+ | Because the only way the product of the two numbers is <math>0</math> is if one of the numbers we choose is <math>0,</math> we calculate the probability of NOT choosing a <math>0.</math> We get <math>\frac{5}{6} \cdot \frac{4}{5} = \frac{2}{3}.</math> Therefore our answer is <math>1 - \frac{2}{3} = \boxed{\textbf{(D)} \ \frac{1}{3}}.</math> | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/cRsvq0BH4MI | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | ==Video Solution by OmegaLearn == | ||
+ | https://youtu.be/6xNkyDgIhEE?t=357 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/jDeS4A6N-nE | ||
− | + | ~savannahsolver | |
− | |||
+ | ==See Also== | ||
{{AMC8 box|year=2016|num-b=12|num-a=14}} | {{AMC8 box|year=2016|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:14, 10 June 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
1. Identify the total number of ways to select two different numbers from the set:
The set has 6 elements. The number of ways to choose 2 different numbers from 6 is given by the combination formula: .
2. Identify the favorable outcomes:
For the product to be zero, one of the chosen numbers must be zero. The set contains one zero (0). To have a product of zero, we need to choose 0 and any other number from the remaining five numbers .
The number of ways to choose 0 and one other number from the remaining five is 5.
3. Calculate the probability:
The probability is the number of favorable outcomes divided by the total number of outcomes: .
Thus, the probability that the product is is
~GeometryMystery
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
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 |
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