Independent solution

How to solve this Geometric Distribution question

Setup

Setup

Count the two qualifying configurations among the eight equally likely outcomes of one experiment.

p=223=14,q=1p=34p=\frac{2}{2^3}=\frac14,\qquad q=1-p=\frac34

Model

Model

A first success on experiment three requires failures on experiments one and two, then success on experiment three.

Pr(T=3)=q2p\Pr(T=3)=q^2p

Compute

Compute

Multiply the independent experiment-level probabilities.

Pr(T=3)=(34)2(14)=964=0.140625\Pr(T=3)=\left(\frac34\right)^2\left(\frac14\right)=\frac9{64}=0.140625

Answer

Answer

The probability rounds to 0.141.

0.141(C)\boxed{0.141\quad\text{(C)}}