Venn diagram, conditional probabilities and correlation

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Consider the Venn diagram for two events $A$ and $B$: the $A$-only region is $0{,}4$, the $B$-only region is $0{,}2$, the intersection is $y$ and the outside is $t$.

• (a) Knowing that $P(A\cap B)=3\,P(\bar A\cap\bar B)$, complete the diagram.
• (b) Compute $P(A\mid B)$ and $P(B\mid A)$.
• (c) Compute the correlation between the two events and $P(A\cup B)$.
• (d) What would $t$, $y$ and the correlation be for incompatible events? And for independent events?

Solution

(a) The four regions sum to $1$: $0{,}4+0{,}2+y+t=1\Rightarrow y+t=0{,}4$. With $y=3t$: $3t+t=0{,}4\Rightarrow t=0{,}1,\quad y=0{,}3.$ So $P(A)=0{,}7$, $P(B)=0{,}5$. (b) $P(A\mid B)=\dfrac{0{,}3}{0{,}5}=0{,}6$; $\;P(B\mid A)=\dfrac{0{,}3}{0{,}7}\approx0{,}43$. (c) Correlation (indicator-variable coefficient): $\rho=\frac{0{,}3-0{,}35}{\sqrt{0{,}7\cdot0{,}3\cdot0{,}5\cdot0{,}5}}\approx-0{,}22.$ $P(A\cup B)=0{,}9$. (d) Incompatible: $y=0$, so $t=0{,}4$; correlation $\rho\approx-0{,}41$. Independent: would need $\rho=0$, i.e. $y=(0{,}4+y)(0{,}2+y)$; the equation $y^2-0{,}4y+0{,}08=0$ has negative discriminant, so with those two side regions independence cannot be achieved. $\boxed{y=0{,}3,\ t=0{,}1,\ P(A\mid B)=0{,}6,\ P(A\cup B)=0{,}9,\ \rho\approx-0{,}22.}$