Among the 640 employees in a company, 60% of males and 50% of females are married. If a randomly selected employee is married, the probability that this person is a male is double the probability that the person is a female. What is the number of female employees in this company?
I am completely lost where I am supposed to start. The only thing I can think of is that this might be a combination problem with relative frequency and product rule involved somehow. I think this is the hardest part for me, not being able to even recognize what type of problem I am dealing with.
I don’t see the connection for where to start with the given fact that 60% of males and 50% of females are married fit into my given hypothesis to use as a combination, relative frequency, or product rule.
Best Answer
Let $A$ be the event that our randomly selected person is married. Let $B$ be the event that our randomly selected person is male.
We are told $Pr(A\mid B)=0.6$ and that $Pr(A\mid B^c)=0.5$
Further, we are told that $Pr(B\mid A)=2Pr(B^c\mid A)$
We know from earlier results that $Pr(B\mid A)+Pr(B^c\mid A)=1$, so this tells us what $Pr(B\mid A)$ is.
Now, let $Pr(B)=x$. We wish to find what $x$ is.
We know that $Pr(A\mid B)=\dfrac{Pr(B\mid A)Pr(A)}{Pr(B)}$ and that $Pr(A)=Pr(B)Pr(A\mid B)+Pr(B^c)Pr(A\mid B^c)$.
Plugging in values and completing the necessary algebra lets us find the value of $x$ from which we can now find the number of females in the company.