Question

A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k = 1,2,3....$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?

$\dfrac{1}{4}$

$\dfrac{2}{7}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\dfrac{3}{7}$

Solution:

Given that the two balls were tossed into separate bins, the probability that the ball in the higher-numbered bin is red is $\frac 12.$ Thus we must find $\frac{P(\text{Balls in different bins})}2$ $= \frac{1-P(\text{Balls in same bins})}2$ by complementary counting.

The probability that both balls are in bin $k$ is $2^{-k} \cdot 2^{-k} = 4^{-k}.$

The probability that they are both in the same bin is therefore $\sum_{k=1}^\infty 4^{-k}.$ Using the geometric sequence formula, we get this to be $\frac 14 \cdot \dfrac{1}{1-\frac 14} = \frac 13.$

Therefore, our answer is $\frac{1-\frac 13}2 = \frac 13 .$

Thus, the answer is C .

2019 AMC 10B Problem 17

Solution:

Problem 17 in Other Years