Q.Events A and B are such that $P(A) = \frac{1}{2}$, $P(B) = \frac{7}{12}$ and P(not A or not B) = $\frac{1}{4}$. State whether A and B are independent ?

Compute $P(E)$, $P(F)$, $P(E\cap F)$ for the two throws and check $P(E\cap F)=P(E)P(F)$.

Sample space for two throws of a die has 36 equally likely outcomes.

$E$ = even number on first throw = $\{2,4,6\}\times\{1,\dots,6\}$, so $|E|=18$, $P(E)=18/36=1/2$.

$F$ = even number on second throw, similarly $P(F)=1/2$.

$E\cap F$ = even on both throws: $3\times3=9$ outcomes, so $P(E\cap F)=9/36=1/4$.

Check independence: $P(E)\cdot P(F)=\dfrac12\cdot\dfrac12=\dfrac14=P(E\cap F)$.

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With $P(A)=\frac12,P(B)=\frac13$ independent: $P(\text{solved})=1-P(A')P(B')=\frac23$; $P(\text{exactly one})=\frac12$. (OR: normalizing the given distribution gives $k=0.15$, from which the requested probabilities follow.)

$P(A)=\dfrac12$, $P(B)=\dfrac13$, independent events

$P(A')=\frac12,\qquad P(B')=1-\frac13=\frac23.$

1. Probability the problem is solved (i.e. at least one of A, B solves it): $P(A\cup B) = 1-P(\text{neither solves}) = 1-P(A')P(B') = 1-\frac12\cdot\frac23 = 1-\frac13=\frac23.$
2. Probability exactly one of them solves it: $P(\text{exactly one}) = P(A)P(B')+P(A')P(B) = \frac12\cdot\frac23 + \frac12\cdot\frac13 = \frac13+\frac16 = \frac12.$

OR: Rita's study-hours distribution

$P(X=0)=0.1,\quad P(X=1)=k,\quad P(X=2)=2k,\quad P(X=3)=2k,\quad P(X=4)=k.$

(since $P(X{=}1)=k(1)=k$, $P(X{=}2)=k(2)=2k$, $P(X{=}3)=k(5-3)=2k$, $P(X{=}4)=k(5-4)=k$.)

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