Venn diagram, conditional probability and correlation

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Consider the Venn diagram for two events $A$ and $B$: the $A$-only region is $0.4$, the intersection $A\cap B$ is $0.1$, the $B$-only region is $0.2$, and the region outside both is $t$ (unknown). (figure in the original test: two overlapping ellipses inside a rectangle)

• (a) compute $t=P(\bar A\cap\bar B)$ and place it in the drawing;
• (b) compute $P(A|B)$;
• (c) compute $P(B|A)$;
• (d) compute $P(B|\bar A)$;
• (e) compute $P(A\cup B)$;
• (f) determine the correlation between the two events.

Solution

The four regions sum to $1$, so $t=1-(0.4+0.1+0.2)=\boxed{0.3}$. Also $P(A)=0.4+0.1=0.5$ and $P(B)=0.1+0.2=0.3$.

(b) $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.1}{0.3}=\dfrac13\approx0.33$. (c) $P(B|A)=\dfrac{0.1}{0.5}=0.2$. (d) $P(\bar A)=0.5$ and $P(B\cap\bar A)=0.2$, so $P(B|\bar A)=\dfrac{0.2}{0.5}=0.4$. (e) $P(A\cup B)=0.4+0.1+0.2=0.7$. (f) Compare $P(A\cap B)$ with $P(A)\,P(B)=0.5\cdot0.3=0.15$. Since $0.1<0.15$, the events are negatively correlated: the occurrence of one makes the other less likely. $\boxed{\ t=0.3;\ \ P(A|B)=\tfrac13;\ \ P(B|A)=0.2;\ \ P(B|\bar A)=0.4;\ \ P(A\cup B)=0.7;\ \text{negative correlation}\ }$