Difference between revisions of "2016 AMC 10B Problems/Problem 12"
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<math>\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8</math> | <math>\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8</math> | ||
− | ==Solution== | + | ==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 <math>\frac{\tbinom32}{\tbinom52}=\frac3{10}</math>, so the answer is <math>1-0.3</math> which is <math>\textbf{(D)}\ 0.7</math>. | + | 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 <math>\frac{\tbinom32}{\tbinom52}=\frac3{10}</math>, so the answer is <math>1-0.3</math> which is <math>\boxed{\textbf{(D) }0.7}</math>. |
+ | |||
+ | An alternate way to finish: | ||
+ | Since it is odd if none are even, the probability is <math>1-(\frac{3}{5} \cdot \frac{2}{4})=1-\frac{3}{10}=0.7 \Longrightarrow \boxed{\textbf{(D) }0.7}</math>. | ||
+ | ~Alternate solve by JH. L | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | 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. | ||
+ | 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>. | ||
+ | |||
+ | Adding these up we get <math>\dfrac{1}{10}+\dfrac{3}{10}+\dfrac{3}{10} = \boxed{\textbf{(D) }0.7}.</math> | ||
+ | |||
+ | -ConfidentKoala4 | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/IRyWOZQMTV8?t=933 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==Video Solution== | ==Video Solution== | ||
− | https://youtu.be/tUpKpGmOwDQ | + | https://youtu.be/tUpKpGmOwDQ - savannahsolver |
− | |||
− | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2016|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 03:35, 4 November 2022
Contents
Problem
Two different numbers are selected at random from and multiplied together. What is the probability that the product is even?
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 , so the answer is which is .
An alternate way to finish: Since it is odd if none are even, the probability is . ~Alternate solve by JH. L
Solution 2
There are cases to get an even number. Case 1: and Case 2: . Thus, to get an , you get . And to get , you get . which is and the answer is .
Solution 3
Note that we have three cases to get an even number: even even, odd even and even odd. The probability of case 1 is , the probability of case 2 is and the probability of case 3 is .
Adding these up we get
-ConfidentKoala4
Video Solution by OmegaLearn
https://youtu.be/IRyWOZQMTV8?t=933
~ pi_is_3.14
Video Solution
https://youtu.be/tUpKpGmOwDQ - savannahsolver
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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