Difference between revisions of "2016 AMC 8 Problems/Problem 21"
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==Solution 1== | ==Solution 1== | ||
| − | We put five chips randomly in order, and then pick the chips from the left to the right. To find the number of ways to rearrange the three red chips and two green chips, we solve for <math>\binom{5}{2} = 10</math>. 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. This means that the last chip must be green in all the situations that work. This means we are left with finding the number of ways to rearrange three red chips and one green chip, which is <math>\binom{4}{3} = 4</math>. Because a green chip will be last <math>4</math> out of the <math>10</math> situations, our answer is | + | We put five chips randomly in order, and then pick the chips from the left to the right. To find the number of ways to rearrange the three red chips and two green chips, we solve for <math>\binom{5}{2} = 10</math>. 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. This means that the last chip must be green in all the situations that work. This means we are left with finding the number of ways to rearrange three red chips and one green chip, which is <math>\binom{4}{3} = 4</math>. Because a green chip will be last <math>4</math> out of the <math>10</math> situations, our answer is <math>\boxed{\textbf{(B) } \frac{2}{5}}</math>. |
==Solution 2== | ==Solution 2== | ||
Revision as of 09:54, 3 November 2019
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 1
We put five chips randomly in order, and then pick the chips from the left to the right. To find the number of ways to rearrange the three red chips and two green chips, we solve for 
. 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. This means that the last chip must be green in all the situations that work. This means we are left with finding the number of ways to rearrange three red chips and one green chip, which is 
. Because a green chip will be last 
out of the 
situations, our answer is 
.
Solution 2
There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out 
red chips, red chips and 
green chip, 
green chips, green chips and red chip, and green chips and red chips. Because order is important in this problem, there are 
ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which 
. Out of the 10 ways to end the game, 4 of them ends with a red chip. The answer is 
, or .
| 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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