Inclusion-Exclusion Probability Question Sample SOA Exam P

Question

You are given $P[A\cup B]=0.7$ and $P[A\cup B^{c}]=0.9$, calculate $P[A]$. No other information is given.

I know $P[A\cup B]=P[A]+P[B]-P[A\cap B]$ and similarly $P[A\cup B^{c}]=P[A]+P[B^{c}]-P[A\cap B^{c}]$.

I'm able to extrapolate from this information that $P[B]-P[A\cap B]=0.1$ from drawing a Venn-diagram example, but I'm not sure how to continue or what other information I need to solve this. I feel I am missing something pretty basic so I appreciate any and all help.

Answer 1

Note that each event in the sample space is either in $B$ or in $B^{C}$, but not both. Thus, you have

$$P[B] + P[B^{C}] = 1 \tag{1}\label{eq1A}$$

Similarly, all events in $A$ are also either in $B$ or in $B^{C}$, but not both, so you get

$$P[A \cap B] + P[A \cap B^{C}] = P[A] \tag{2}\label{eq2A}$$

Adding your two expressions for unions of sets, and gathering appropriate terms, gives

$$\begin{equation}\begin{aligned} & P[A\cup B] + P[A\cup B^{c}] \\ & = (P[A]+P[B]-P[A\cap B]) + (P[A]+P[B^{c}]-P[A\cap B^{c}]) \\ & = 2P[A] + (P[B] + P[B^{C}]) - (P[A \cap B] + P[A \cap B^{C}]) \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Finally, using the given values, as well as \eqref{eq1A} and \eqref{eq2A}, results in

$$\begin{equation}\begin{aligned} 0.7 + 0.9 & = 2P[A] + 1 - P[A] \\ 1.6 & = P[A] + 1 \\ P[A] & = 0.6 \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Answer 2

Alternatively, $$\begin{align*} \Pr[A] &= \Pr[A \cup B] + \Pr[A \cap B] - \Pr[B] \\ &= \Pr[A \cup B] - \Pr[A^c \cap B] \\ &= \Pr[A \cup B] - \Pr[(A \cup B^c)^c] \\ &= \Pr[A \cup B] - (1 - \Pr[A \cup B^c]) \\ &= 0.6.\end{align*} $$ This line of reasoning is easily discovered by drawing the Venn diagram.