probability hypergeometric distribution An HR manager estimates that 35% of married employees in a large office complex have spouses whose employers provide dental insurance and 65% have spouses whose employers provide extended medical and drug insurance. Of those whose spouses have dental insurance, 90% have also extended medical and drug insurance.
a) What is the probability that a randomly selected married employee has a spouse who has both types of insurance?
Given:
Dental = 35%
Medical and Drug = 65%
Dental, medical and drug = 90% of medical and drug = 58.5%
So therefore, medical and drug ONLY = 6.5%
So the probability of (a) would be 58.5%?
b) What is the probability that a married employee has a spouse who has neither type of insurance?
Would this be ZERO? Since there is no mention of employees whose spouses are without insurance. Rather it states that all married employees have spouses who have SOME type of insurance.
I can't seem to use the nCr theorem when only percentages are given with no fixed range. I have been stuck on this since the past 60 minutes and I don't know where to start from. How can I use the nCr probability distribution theorem for this particular type of a question? Support would be greatly appreciated. 
 A: Let $D$ be the event "spouse has dental insurance" and $E$ be the event "spouse has extended medical." The two events overlap: some spouses have both.
Note that $\Pr(D)=0.35$ and $\Pr(E)=0.65$. (These add up to $1$. That's an "accident," and perhaps deliberately chosen to increase confusion. Not nice!)
We are told that $90\%$ of the people in $D$ are also in $E$. So the probability of both is $(0.35)(0.9)$. That's the first answer.
For the second question, we want to find the probability of landing outside both $D$ and $E$.
Note that if we add $0.35$ and $0.65$, we have counted the people who are in both twice. so the probability that a person is in both $D$ and $E$ is
$$0.35+0.65-(0.35)(0.9).\tag{1}$$
The probability of landing outside both $D$ and $E$ is therefore $1$ minus the number in (1). This is
$$1-\left(0.35+0.65-(0.35)(0.9)  \right).$$
This happens to simplify considerably.
*Remark: We are told something that lets us find that $\Pr(D\cap E)=(0.35)(0.9)$. That's the answer to the first question.
Now use the formula
$$\Pr(D\cup E)=\Pr(D)+\Pr(E)-\Pr(D\cap E)$$
to calculate $\Pr(D\cup E)$. For the second question, we want the probability of neither $D$ nor $E$, that is, the probability of the complement of $D\cup E$. This is $1-\Pr(D\cup E)$.
