Difference between revisions of "2006 AMC 10B Problems/Problem 17"
Mathkiddie (talk | contribs) (This solution is unnessasary, as it is nearly the same as Solution 1.) |
Pi is 3.14 (talk | contribs) (→Video Solution) |
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<math> \textbf{(A) } \frac{1}{10}\qquad \textbf{(B) } \frac{1}{6}\qquad \textbf{(C) } \frac{1}{5}\qquad \textbf{(D) } \frac{1}{3}\qquad \textbf{(E) } \frac{1}{2} </math> | <math> \textbf{(A) } \frac{1}{10}\qquad \textbf{(B) } \frac{1}{6}\qquad \textbf{(C) } \frac{1}{5}\qquad \textbf{(D) } \frac{1}{3}\qquad \textbf{(E) } \frac{1}{2} </math> | ||
| − | == Video Solution == | + | == Video Solution by OmegaLearn == |
https://youtu.be/5UojVH4Cqqs?t=1160 | https://youtu.be/5UojVH4Cqqs?t=1160 | ||
Latest revision as of 03:32, 4 November 2022
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?
Video Solution by OmegaLearn
https://youtu.be/5UojVH4Cqqs?t=1160
~ pi_is_3.14
Solution
Since there are the same amount of total balls in Alice's bag as in 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 
red balls out of the 
balls in his bag.
So the desired 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. 
