Difference between revisions of "2016 AMC 8 Problems/Problem 21"
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==Solution== | ==Solution== | ||
| − | We put five chips randomly in order, and then pick the chips from the left to the right. However, we notice that whenever the last chip we draw is red, we pick both green chips before we pick the last (red) chip. Similarly, when the last chip is green, we pick all three red chips before the last (green) chip. | + | We put five chips randomly in order, and then pick the chips from the left to the right. However, we notice that whenever the last chip we draw is red, we pick both green chips before we pick the last (red) chip. Similarly, when the last chip is green, we pick all three red chips before the last (green) chip. Because a green chip will be last 6 out of 10 times and a red chip will be last 4 out of 10 times, our answer is <math>\boxed{\text{(B) }\dfrac{2}{5}}</math>. |
{{AMC8 box|year=2016|num-b=20|num-a=22}} | {{AMC8 box|year=2016|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:44, 23 November 2016
A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?
Solution
We put five chips randomly in order, and then pick the chips from the left to the right. However, we notice that whenever the last chip we draw is red, we pick both green chips before we pick the last (red) chip. Similarly, when the last chip is green, we pick all three red chips before the last (green) chip. Because a green chip will be last 6 out of 10 times and a red chip will be last 4 out of 10 times, our answer is 
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| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
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