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. 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 4 out of 10 times and a red chip will be last 6 out of 10 times, our answer is <math>\boxed{\ | + | 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 <math>4</math> out of <math>10</math> times and a red chip will be last <math>6</math> out of <math>10</math> times, our answer is <math>\boxed{\textbf{(B) } \frac{2}{5}}</math>. |
==Solution 2== | ==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 3 red chips, 3 red chips and | + | 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 <math>3</math> red chips, <math>3</math> red chips and <math>1</math> green chip, <math>2</math> green chips, <math>2</math> green chips and <math>1</math> red chip, and <math>2</math> green chips and <math>2</math> red chips. Because order is important in this problem, there are <math>1+4+1+3+6=15</math> 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 <math>15-5=10</math>. Out of the 10 ways to end the game, 4 of them ends with a red chip. The answer is <math>\frac{4}{10} = \frac{2}{5}</math>, or <math>\boxed{\textbf{(B) } \frac{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 00:49, 28 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 1
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 out of
times and a red chip will be last
out of
times, 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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