Independent solution

How to solve this Geometric Distribution question

Answer in brief

This is a geometric stopping-time calculation. A round continues with probability (1/4)^3+(3/4)^3=7/16, so exactly five rounds requires four continuations followed by a stop: (7/16)^4(9/16)=0.020608, making choice C correct.

Setup

Setup

In one round, continuation occurs only when all three outcomes match. Add the all-first-outcome and all-second-outcome probabilities.

r=(14)3+(34)3=716r=\left(\frac14\right)^3+\left(\frac34\right)^3=\frac7{16}

Model

Model

Exactly five rounds means that each of the first four rounds continues and the fifth round stops.

Pr⁡(T=5)=r4(1−r)\Pr(T=5)=r^4(1-r)

Compute

Compute

Substitute the one-round continuation and stopping probabilities.

Pr⁡(T=5)=(716)4(916)\Pr(T=5)=\left(\frac7{16}\right)^4\left(\frac9{16}\right)
Pr⁡(T=5)=216091048576=0.0206079483…\Pr(T=5)=\frac{21609}{1048576}=0.0206079483\ldots

Answer

Answer

The probability rounds to 0.0206.

0.0206(C)\boxed{0.0206\quad\text{(C)}}