Independent solution

How to solve this Law of Total Probability question

Setup

Setup

Let h be the probability of selecting the weighted die. Normalizing its face weights gives probability 1/9 on each odd face and 2/9 on each even face.

h=Pr⁡(weighted-die branch)h=\Pr(\text{weighted-die branch})
Pr⁡(X=k∣weighted)={1/9,k odd,2/9,k even.\Pr(X=k\mid\text{weighted})=\begin{cases}1/9,&k\text{ odd},\\2/9,&k\text{ even}.\end{cases}

Model

Model

Apply the law of total probability to the event that the displayed value is at most three.

Pr⁡(X≤3∣weighted)=1+2+19=49\Pr(X\le3\mid\text{weighted})=\frac{1+2+1}{9}=\frac49
Pr⁡(X≤3∣fair)=12\Pr(X\le3\mid\text{fair})=\frac12
0.463=h(49)+(1−h)(12)0.463=h\left(\frac49\right)+(1-h)\left(\frac12\right)

Compute

Compute

Solve for the mixing probability, then mix the two conditional probabilities for face four.

h=18(0.500−0.463)=0.666h=18(0.500-0.463)=0.666
Pr⁡(X=4)=0.666(29)+0.334(16)\Pr(X=4)=0.666\left(\frac29\right)+0.334\left(\frac16\right)
Pr⁡(X=4)=0.203666…\Pr(X=4)=0.203666\ldots

Answer

Answer

The requested probability rounds to 0.204 at the precision of the choices.

Pr⁡(X=4)≈0.204(C)\boxed{\Pr(X=4)\approx0.204\quad\text{(C)}}