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Difference between revisions of "2016 AMC 10B Problems/Problem 12"

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==Solution 2==
 
==Solution 2==
 
There are <math>2</math> cases to get an even number. Case 1: <math>\text{Even} \times \text{Even}</math> and Case 2: <math>\text{Odd} \times \text{Even}</math>. Thus, to get an <math>\text{Even} \times \text{Even}</math>, you get <math>\frac {\binom {2}{2}}{\binom {5}{2}}= \frac {1}{10}</math>. And to get <math>\text{Odd} \times \text{Even}</math>, you get <math>\frac {\binom {3}{1}}{\binom {5}{2}}= \frac {6}{10}</math>. <math>\frac {1}{10}+\frac {6}{10}=\frac {7}{10}</math> which is <math>0.7</math> and the answer is <math>\boxed{\textbf{(D) }0.7}</math>.
 
There are <math>2</math> cases to get an even number. Case 1: <math>\text{Even} \times \text{Even}</math> and Case 2: <math>\text{Odd} \times \text{Even}</math>. Thus, to get an <math>\text{Even} \times \text{Even}</math>, you get <math>\frac {\binom {2}{2}}{\binom {5}{2}}= \frac {1}{10}</math>. And to get <math>\text{Odd} \times \text{Even}</math>, you get <math>\frac {\binom {3}{1}}{\binom {5}{2}}= \frac {6}{10}</math>. <math>\frac {1}{10}+\frac {6}{10}=\frac {7}{10}</math> which is <math>0.7</math> and the answer is <math>\boxed{\textbf{(D) }0.7}</math>.
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==Solution 3==
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Note that we have three cases to get an even number: even <math>\times</math> even, odd <math>\times</math> even and even <math>\times</math> odd.
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The probability of case 1 is <math>\dfrac{2}{5}\cdot\dfrac{1}{4}</math>, the probability of case 2 is <math>\dfrac{2}{5}\cdot\dfrac{3}{4}</math> and the probability of case 3 is <math>\dfrac{3}{5}\cdot\dfrac12</math>.
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Adding these up we get <math>\dfrac{1}{10}+\dfrac{3}{10}+\dfrac{3}{10} = \boxed{\textbf{(D) }0.7}.</math>
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-ConfidentKoala4
  
 
==Video Solution==
 
==Video Solution==

Revision as of 17:45, 31 August 2022

Problem

Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?

$\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8$

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 $\frac{\tbinom32}{\tbinom52}=\frac3{10}$, so the answer is $1-0.3$ which is $\boxed{\textbf{(D) }0.7}$.

An alternate way to finish: Since it is odd if none are even, the probability is $1-(\frac{3}{5} \cdot \frac{2}{4})=1-\frac{3}{10}=0.7 \Longrightarrow \boxed{\textbf{(D) }0.7}$. ~Alternate solve by JH. L

Solution 2

There are $2$ cases to get an even number. Case 1: $\text{Even} \times \text{Even}$ and Case 2: $\text{Odd} \times \text{Even}$. Thus, to get an $\text{Even} \times \text{Even}$, you get $\frac {\binom {2}{2}}{\binom {5}{2}}= \frac {1}{10}$. And to get $\text{Odd} \times \text{Even}$, you get $\frac {\binom {3}{1}}{\binom {5}{2}}= \frac {6}{10}$. $\frac {1}{10}+\frac {6}{10}=\frac {7}{10}$ which is $0.7$ and the answer is $\boxed{\textbf{(D) }0.7}$.


Solution 3

Note that we have three cases to get an even number: even $\times$ even, odd $\times$ even and even $\times$ odd. The probability of case 1 is $\dfrac{2}{5}\cdot\dfrac{1}{4}$, the probability of case 2 is $\dfrac{2}{5}\cdot\dfrac{3}{4}$ and the probability of case 3 is $\dfrac{3}{5}\cdot\dfrac12$.

Adding these up we get $\dfrac{1}{10}+\dfrac{3}{10}+\dfrac{3}{10} = \boxed{\textbf{(D) }0.7}.$

-ConfidentKoala4

Video Solution

https://youtu.be/tUpKpGmOwDQ - savannahsolver

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
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 AMC 10 Problems and Solutions

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