Difference between revisions of "2016 AMC 8 Problems/Problem 21"

(Solution 1)
 
(2 intermediate revisions by the same user not shown)
Line 8: Line 8:
 
==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 <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>.
 
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>.
 +
 +
==Solution 3==
 +
Assume that after you draw the three red chips in a row without drawing both green chips, you continue drawing for the next turn. The last/fifth chip that is drawn must be a green chip. So technically, the problem is asking for the probability that the "fifth draw" is a green chip. This probability is symmetric to the probability that the first chip drawn is green, which is <math>\frac{2}{5}</math>. Thus the probability is <math>\boxed{\textbf{(B) } \frac{2}{5}}</math>. ~skyscraper
 +
 +
Note: This problem is almost identical to 2001 AMC 10 #23.
  
 
{{AMC8 box|year=2016|num-b=20|num-a=22}}
 
{{AMC8 box|year=2016|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:42, 10 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?

$\textbf{(A) }\dfrac{3}{10}\qquad\textbf{(B) }\dfrac{2}{5}\qquad\textbf{(C) }\dfrac{1}{2}\qquad\textbf{(D) }\dfrac{3}{5}\qquad \textbf{(E) }\dfrac{7}{10}$

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 $\binom{5}{2} = 10$. 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 $\binom{4}{3} = 4$. Because a green chip will be last $4$ out of the $10$ situations, our answer is $\boxed{\textbf{(B) } \frac{2}{5}}$.

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 $1$ green chip, $2$ green chips, $2$ green chips and $1$ red chip, and $2$ green chips and $2$ red chips. Because order is important in this problem, there are $1+4+1+3+6=15$ 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 $15-5=10$. Out of the 10 ways to end the game, 4 of them ends with a red chip. The answer is $\frac{4}{10} = \frac{2}{5}$, or $\boxed{\textbf{(B) } \frac{2}{5}}$.

Solution 3

Assume that after you draw the three red chips in a row without drawing both green chips, you continue drawing for the next turn. The last/fifth chip that is drawn must be a green chip. So technically, the problem is asking for the probability that the "fifth draw" is a green chip. This probability is symmetric to the probability that the first chip drawn is green, which is $\frac{2}{5}$. Thus the probability is $\boxed{\textbf{(B) } \frac{2}{5}}$. ~skyscraper

Note: This problem is almost identical to 2001 AMC 10 #23.

2016 AMC 8 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS