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

Revision as of 09:19, 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 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 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 $\frac{2}{5}$ . So the probability is $\boxed{\textbf{(B) } \frac{2}{5}}$ .

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

Solution 5

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

@@ Line 13: / Line 13: @@
 Note: This problem is almost identical to 2001 AMC 10 Problem #23.
-===Solution 4===
-If we list all possibilities where the <math>3</math> red chips are drawn , where R represents Red and G represents Green, they are RRR, GRRR, RGRR, and RRGR, because the last chip drawn must be red. This means that there are 4 possibilities for it to be red. The only other outcome is that it ends in green, which if we represent them the same way we did for red, we get, GG, GRG, RGG, RRGG, RGRG, and GRRG, or <math>6</math> outcomes. This means there are <math>4+6=10</math> possible outcomes, and 4 outcomes that result in <math>3</math> red being drawn, so the probability is <math>\frac{4}{10} = \frac{2}{5}</math>, and the answer is <math>\boxed{\textbf{(B) } \frac{2}{5}}</math>
-~~DaLegendaryTaco72
 ===Solution 5===

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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