Conditional probability with hidden dependence

Question

There are an equal number of male and female workers in a company. In any given year, the probability that a male gets a lucky bonus is $\alpha$, independently of other years. the same probability for females is $\beta$. Assume the company randomly selects a worker.

let event $A_1$, $A_2$ denote respectively that a randomly chosen worker gets a bonus in each of the first and second years.

I know that $P(A_2|A_1) \geq P(A_1)$. Intuitively, if $\alpha$ and $\beta$ are unequal to each other, we can get some information from the result of the first year.

I saw the answer is $P(A_2|A_1)-P(A_1)=\frac{(\alpha-\beta)^2}{2(\alpha+\beta)}$. But I don't know how to get there since $A_1$ $A_2$ are not independent.

Answer 1

Hints: $M$ is the event the selected person is male. $F$ female. What we know is actually some conditional probabilities, $P(A_1|M)=\alpha$, $P(A_1|F)=\beta$.

From Beyes's rule, we can get $P(M|A_1)=\frac{P(A_1|M)P(M)}{P(A_1)}=\frac{\alpha/2}{P(A_1)}$ where $P(A_1)=\frac{\alpha+\beta}{2}$. Similarly, you can get $P(F|A_1).$

To get $P(A_2|A_1)$, you can think of the expression $P(A_2|A_1)=P(A_2|M)P(M|A_1)+P(A_2|F)P(F|A_1)$. What does it mean in the scenario described?

Answer 2

Use the law of total probability. Let $M$ be the event of a male worker and $F$ be the event of a female worker. Let $A_{1}$ be the event of a bonus in the first year and $A_2$ be the even of a bonus in the second year. Then $$ P(A_{1})=P(A_{1}\mid M)P(M)+P(A_{1}\mid F)P(F)=\alpha\times\frac{1}{2}+\beta\times\frac{1}{2}=\frac{\alpha+\beta}{2} $$ since there are equal number of male and female workers and the worker is chosen randomly. Similarly $$ \begin{align} P(A_{1}\cap A_{2})&=P(A_{1}\cap A_{2})\mid M)P(M)+P(A_{1}\cap A_{2})\mid F)P(F)\\ &=\frac{\alpha^2}{2}+\frac{\beta^2}{2} \end{align} $$ since lucky bonuses are independent events conditional on being male or female. Then $$ P(A_{2}\mid A_{1})=\frac{P(A_1\cap A_2)}{P(A_{1})}=\frac{\alpha^2+\beta^2}{\alpha+\beta} $$ whence $$ P(A_{2}\mid A_{1})-P(A_1)=\frac{(\alpha-\beta)^2}{2(\alpha+\beta)}. $$