Difference between revisions of "2006 AMC 10B Problems/Problem 17"
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This has a <math>\frac{2}{6} = \frac{1}{3}</math> chance. | This has a <math>\frac{2}{6} = \frac{1}{3}</math> chance. | ||
| − | Since Alice can choose <math>5</math> total colours and the probability for each is the same since there are the same number of balls for each, the probability is <math>\frac{1}{5} \cdot \frac{1}{3} \cdot 5 = \boxed{\ | + | Since Alice can choose <math>5</math> total colours and the probability for each is the same since there are the same number of balls for each, the probability is <math>\frac{1}{5} \cdot \frac{1}{3} \cdot 5 = \boxed{\textbf{(D)}\frac{1}{3}}</math> |
== See Also == | == See Also == | ||
Revision as of 21:45, 30 December 2020
Contents
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?
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 
Solution 2
Suppose Alice selects the blue ball. This happens in a 
chance.
Then Bob has to select blue, too and now he has 
blue balls and 
total marbles.
This has a 
chance.
Since Alice can choose 
total colours and the probability for each is the same since there are the same number of balls for each, the probability is 
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
