Difference between revisions of "2006 AMC 10B Problems/Problem 17"

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<math> \mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{5}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math>
 
<math> \mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{5}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math>
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== Video Solution ==
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https://youtu.be/5UojVH4Cqqs?t=1160
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~ pi_is_3.14
  
 
== Solution ==
 
== Solution ==

Latest revision as of 21:33, 24 January 2021

Problem

Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same?

$\mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{5}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2}$

Video Solution

https://youtu.be/5UojVH4Cqqs?t=1160

~ pi_is_3.14

Solution

Since there are the same amounts of all the balls in Alice's bag and Bob's bag, and there is an equal chance of each ball being selected, the color of the ball that Alice puts in Bob's bag doesn't matter. Without loss of generality, let the ball Alice puts in Bob's bag be red.

For both bags to have the same contents, Bob must select one of the 2 red balls out of the 6 balls in his bag.

So the desired probability is $\frac{2}{6} = \frac{1}{3} \Rightarrow D$

Solution 2

Suppose Alice selects the blue ball. This happens in a $\frac{1}{5}$ chance.

Then Bob has to select blue, too and now he has $2$ blue balls and $6$ total marbles.

This has a $\frac{2}{6} = \frac{1}{3}$ chance.

Since Alice can choose $5$ total colours and the probability for each is the same since there are the same number of balls for each, the probability is $\frac{1}{5} \cdot \frac{1}{3} \cdot 5 = \boxed{\textbf{(D)}\frac{1}{3}}$

~mathboy282

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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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