Difference between revisions of "2016 AMC 8 Problems/Problem 13"

Latest revision as of 21:14, 10 June 2024

Problem

Two different numbers are randomly selected from the set $\{ - 2, -1, 0, 3, 4, 5\}$ and multiplied together. What is the probability that the product is $0$ ?

$\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}$

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: $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$ .

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 $-2, -1, 3, 4, 5$ .

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: $\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{15} = \frac{1}{3}$ .

Thus, the probability that the product is $0$ is $\boxed{\textbf{(D)} \ \frac{1}{3}}.$

~GeometryMystery

Solution 2 (Complementary Counting)

Because the only way the product of the two numbers is $0$ is if one of the numbers we choose is $0,$ we calculate the probability of NOT choosing a $0.$ We get $\frac{5}{6} \cdot \frac{4}{5} = \frac{2}{3}.$ Therefore our answer is $1 - \frac{2}{3} = \boxed{\textbf{(D)} \ \frac{1}{3}}.$

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

@@ Line 8: / Line 8: @@
 ===Solution 1===
-The product can only be <math>0</math> if one of the numbers is <math>0</math>. Once we chose <math>0</math>, there are <math>5</math> ways we can chose the second number, or <math>6-1</math>. There are <math>\dbinom{6}{2}</math> ways we can chose <math>2</math> numbers randomly, and that is <math>15</math>. So, <math>\frac{5}{15}=\frac{1}{3}</math> so the answer is  <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math>.
+. Identify the total number of ways to select two different numbers from the set:
-===Solution 2 (Complementary Counting)===
+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>.
-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>
+. Identify the favorable outcomes:
-===Solution 3 (Casework)===
+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>.
-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: <math>0</math> is the first number chosen
+The number of ways to choose 0 and one other number from the remaining five is 5.
-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.
+. 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>.
-Case 2: <math>0</math> is the second number chosen
+Thus, the probability that the product is <math>0</math> is <math>\boxed{\textbf{(D)} \ \frac{1}{3}}.</math>
-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.
+~GeometryMystery
-Adding the probabilities from the two distinct cases up, we find that there is a <math>\frac{1}{6} + \frac{1}{6} = \boxed{\textbf{(D)} \ \frac{1}{3}}</math> chance of getting a product of zero.
+===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!!!)==

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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