Law of Total Probability with Conditionals

Question

I would greatly appreciate some help on this problem, thank you! It is a past problem, so an answer is acceptable with steps.

Suppose we have a sample space $S$ and two events $A$ and $B$ such that

$$\begin{array}{lcl}S=A\cup B & \text{and} & A\cap B=\emptyset\end{array}$$

and an event $E$,

Suppose we have the following results:

• $P\left(A\right)=1/3$
• $P\left(B\right)=2/3$
• $P\left(E\mid A\right)=1/2$
• $P\left(E\mid B\right)=1/4$

What is $P\left(E\right)\thinspace$?

This is all I know:

$\begin{array}{rcl} P\left(E\mid A\right) & = & \frac{P\left(A\cap E\right)}{P\left(A\right)}\\ & =\\ & = & 1/2 \end{array}$

$\begin{array}{rcl} P\left(E\mid B\right) & = & \frac{P\left(B\cap E\right)}{P\left(B\right)}\\ & = & 1/4 \end{array}$

EDIT:

I figured it out. I was completely overthinking it.

$\begin{array}{rcccl} P\left(E\mid A\right) & = & 1/2 & = & \frac{P\left(A\cap E\right)}{P\left(A\right)}\\ & = & & = & \frac{P\left(A\cap E\right)}{1/3}\\ & = & P\left(A\cap E\right) & = & 1/6 \end{array}$

$\begin{array}{rcccl} P\left(E\mid B\right) & = & 1/4 & = & \frac{P\left(B\cap E\right)}{P\left(B\right)}\\ & = & & = & \frac{P\left(B\cap E\right)}{2/3}\\ & & P\left(B\cap E\right) & = & 1/6 \end{array}$

Using the law of total probability,

$P\left(E\right)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$

Answer 1

Since $S$ is the sample space, then use what you have been told about $S,A,B$ to express this in terms of the provided probabilities.

$$\begin{align}\mathsf P(E)&=\mathsf P(S\cap E)\\&=\mathsf P((A\cup B)\cap E)\\&~~\vdots\end{align}$$

$$\begin{align}\mathsf P(E)&=\mathsf P(S\cap E)\\&=\mathsf P((A\cup B)\cap E)\\&=\mathsf P((A\cap E)\cup(B\cap E))\\&=\mathsf P(A\cap E)+\mathsf P(B\cap E)\\&=\mathsf P(A)\mathsf P(E\mid A)+\mathsf P(B)\mathsf P(E\mid B)\end{align}$$

Answer 2

We know that $S = A \cup B$ and $A \cap B = \emptyset$. So $A, B$ are disjoint events whose union is the entire sample space.

Hence we can apply the law of total probability:

$$P(E) = P(A)P(E \mid A) + P(B)P(E \mid B)$$

and all the probabilities on the RHS are known.