Question

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?

$\dfrac{3}{10}$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

Answer: B

Concepts:basic probability symmetry

Difficulty rating: 1490

Video solution:

Click to load, then click again to play

Written solution:

The $3$ reds are drawn before both green chips exactly when a green chip is the last chip in the full ordering. There are $10$ equally likely ways to choose the two positions of the green chips, and $4$ of them have a green chip in the last position.

Therefore, the desired probability is $\dfrac{4}{10} = \dfrac{2}{5}.$

Thus, B is the correct answer.

2016 AMC 8 Problem 21

Answer: B

Difficulty rating: 1490

Video solution:

Written solution:

Problem 21 in Other Years