Conditional probability with a Venn diagram (alligator)

Text

In the Venn diagram, event $A$ is “it rains tomorrow”, event $B$ is “an alligator crosses the square”. The region probabilities are: $A$-only $=0.4$, $B$-only $=0.2$, intersection $A\cap B=y$, outside $=t$.

• (a) In Carpi $P(B\mid A)=20\%$. Find $y$ and $t$.
• (b) In Johannesburg instead $y=0.3$ and $t=0.1$: compute $P(B\mid A)$ and $P(B\mid \neg A)$; decide whether the events are positively correlated, negatively correlated or independent.
• (c) (Johannesburg) Compute the probability that tomorrow it does not rain or an alligator does not cross the square.
• (d) (Johannesburg) Compute the probability that it does not rain given that an alligator does not cross the square.

Solution

(a) $P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{y}{0.4+y}=0.2\Rightarrow y=0.08+0.2y\Rightarrow 0.8y=0.08\Rightarrow y=0.1$. Since everything sums to $1$: $0.4+0.2+0.1+t=1\Rightarrow t=0.3$.

(b) With $y=0.3,\ t=0.1$: $P(A)=0.4+0.3=0.7$, $P(B)=0.2+0.3=0.5$. $P(B\mid A)=\frac{0.3}{0.7}=\frac37\approx 0.429,\qquad P(B\mid\neg A)=\frac{0.2}{0.2+0.1}=\frac{0.2}{0.3}=\frac23\approx 0.667.$ Since $P(B\mid A)=0.429<P(B)=0.5$ (equivalently $P(A\cap B)=0.3<P(A)P(B)=0.35$), the events are negatively correlated.

(c) By De Morgan $\neg A\cup\neg B=\neg(A\cap B)$, so $P(\neg A\cup\neg B)=1-P(A\cap B)=1-0.3=0.7$.

(d) $\neg B$ has probability $0.4+0.1=0.5$; $\neg A\cap\neg B$ is the outside $=0.1$: $P(\neg A\mid\neg B)=\frac{0.1}{0.5}=0.2.$

$\boxed{y=0.1,\ t=0.3;\ P(B\mid A)=\tfrac37,\ P(B\mid\neg A)=\tfrac23\ (\text{neg. corr.});\ P(\neg A\cup\neg B)=0.7;\ P(\neg A\mid\neg B)=0.2}$