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Difference between revisions of "2020 AMC 10B Problems/Problem 11"

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==Solution 2==
 
==Solution 2==
  
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We can analyze this as two containers with <math>10</math> balls each, with the two people grabbing <math>5</math> balls each. First, we need to find the probability of two of the balls being the same among five: <math>\frac{1}{3}\frac{4}{9}\frac{3}{8}\frac{5}{7}\frac{2}{4}</math>. After that we must, multiply this probability by 5 choose 2, for the 2 balls that are the same are chosen among 5 balls. The answer will be 5/126*10 = 25/63. D) 25/63
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We can analyze this as two containers with <math>10</math> balls each, with the two people grabbing <math>5</math> balls each. First, we need to find the probability of two of the balls being the same among five: <math>\frac{1}{3}*\frac{4}{9}*\frac{3}{8}*\frac{5}{7}*\frac{2}{4}</math>. After that we must, multiply this probability by 5 choose 2, for the 2 balls that are the same are chosen among 5 balls. The answer will be 5/126*10 = 25/63. D) 25/63
  
 
==Video Solution==
 
==Video Solution==

Revision as of 11:53, 19 May 2020

Problem

Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?

$\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}$

Solution

We don't care about which books Harold selects. We just care that Betty picks $2$ books from Harold's list and $3$ that aren't on Harold's list.

The total amount of combinations of books that Betty can select is $\binom{10}{5}=252$.

There are $\binom{5}{2}=10$ ways for Betty to choose $2$ of the books that are on Harold's list.

From the remaining $5$ books that aren't on Harold's list, there are $\binom{5}{3}=10$ ways to choose $3$ of them.

$\frac{10\cdot10}{252}=\boxed{\textbf{(D) }\frac{25}{63}}$ ~quacker88

Solution 2

We can analyze this as two containers with $10$ balls each, with the two people grabbing $5$ balls each. First, we need to find the probability of two of the balls being the same among five: $\frac{1}{3}*\frac{4}{9}*\frac{3}{8}*\frac{5}{7}*\frac{2}{4}$. After that we must, multiply this probability by 5 choose 2, for the 2 balls that are the same are chosen among 5 balls. The answer will be 5/126*10 = 25/63. D) 25/63

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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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