Question

Given that $P\left(AUB\right)=0.76$ and $P(AUB\text{'})=0.87$, find $P\left(A\right)$:

Answer 1

Find the probability of event $A$:

Given: $P\left(AUB\right)=0.76$ and $P(AUB\text{'})=0.87$

We know that $\left(A\cup B\right)\cap \left(A\cup B\text{'}\right)=A$ .

So,$P\left((A\cup B)\cap (A\cup B\text{'})\right)=P\left(A\right)$
Recall the inclusion/exclusion principle:

$P\left((A\cup B)\cap (A\cup B\text{'})\right)=P(A\cup B)+P(A\cup B\text{'})-P\left((A\cup B)\cup (A\cup B\text{'})\right)$
As $\left(A\cup B\right)\cup \left(A\cup B\text{'}\right)=U$ (the universal set), we have

$$\begin{gathered} P\left(A\right)=P(A\cup B)+P(A\cup B\text{'})-P\left(U\right) \\ P\left(A\right)=0.76+0.87-1 \\ P\left(A\right)=0.63 \end{gathered}$$

Hence, $P\left(A\right)=0.63$