Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
m (→‎Solution 2)
 
(30 intermediate revisions by 11 users not shown)
Line 7: Line 7:
 
==Solutions==
 
==Solutions==
 
===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 <math>\boxed{\textbf{(B) } \frac{2}{5}}</math>.
+
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===
−
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 green chip. The answer is <math>\frac{4}{10} = \boxed{\textbf{(B) } \frac{2}{5}}</math>.
+
We can use basic probability to find the answer to this question. The probability of the first chip being red is <math>\frac{3}{5}</math> because there are <math>3</math> red chips and <math>5</math> chips in total. The probability of the second chip being red is <math>\frac{2}{4}</math> because there are now <math>2</math> red chips and <math>4</math> total chips. The probability of the third chip being red is <math>\frac{1}{3}</math> because there are now <math>1</math> red chips and <math>3</math> total chips. If we multiply these fractions, we get <math>\frac{1}{10}</math>. To get the answer of <math>\frac{2}{5}</math> we need to multiply this product by <math>4</math> because there are <math>4</math> different draws that these red chips can be chosen on. They can be chosen on the <math>\text{1st 2nd, 3rd, or 4th}</math> draws because if a red chip was chosen on the <math>\text{5th}</math> draw, <math>2</math> green chips would have already been chosen. So our answer is <math>\frac{1}{10}</math> multiplied by <math>4</math> which would leave us with the answer of <math>\boxed{\textbf{(B) } \frac{2}{5}}</math>.
  
−
===Solution 3===
+
~ThatCarGuy3
−
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 because if both green chips were drawn before, we would've already completed the game. 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>. So the probability is <math>\boxed{\textbf{(B) } \frac{2}{5}}</math>.
 
−
 
 
−
Note: This problem is almost identical to 2001 AMC 10 Problem #23.
 
  
 
==Video Solution==
 
==Video Solution==
 +
https://youtu.be/RK1lG84nr4w
  
−
https://youtu.be/fTtUAtfWKyQ - Happytwin
+
~Education, the Study of Everything
−
 
 
−
https://www.youtube.com/watch?v=w0y-JuRQvDc&feature=youtu.be
 
−
 
 
−
https://youtu.be/OOdK-nOzaII?t=1452
 
−
 
 
−
https://youtu.be/m834gDVyPJM
 
−
 
 
−
https://youtu.be/qDIMz0Fd56Q
 
−
 
 
−
~savannahsolver
 
  
 
==See Also==
 
==See Also==

Latest revision as of 13:48, 24 July 2024

Problem

A top hat 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?

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

Solutions

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

We can use basic probability to find the answer to this question. The probability of the first chip being red is $\frac{3}{5}$ because there are $3$ red chips and $5$ chips in total. The probability of the second chip being red is $\frac{2}{4}$ because there are now $2$ red chips and $4$ total chips. The probability of the third chip being red is $\frac{1}{3}$ because there are now $1$ red chips and $3$ total chips. If we multiply these fractions, we get $\frac{1}{10}$. To get the answer of $\frac{2}{5}$ we need to multiply this product by $4$ because there are $4$ different draws that these red chips can be chosen on. They can be chosen on the $\text{1st 2nd, 3rd, or 4th}$ draws because if a red chip was chosen on the $\text{5th}$ draw, $2$ green chips would have already been chosen. So our answer is $\frac{1}{10}$ multiplied by $4$ which would leave us with the answer of $\boxed{\textbf{(B) } \frac{2}{5}}$.

~ThatCarGuy3

Video Solution

https://youtu.be/RK1lG84nr4w

~Education, the Study of Everything

See Also

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_21&oldid=223540"