Probability of simultaneously selecting multiple objects from a set can you please show me how I can solve the following?
A house contains 11 male residents and 28 female residents. Two of the residents are simultaneously selected at random and are evicted from the house. What is the probability that one is male and the other female?
 A: Hint: The required probability equals the number of ways in which we can select 1 male and 1 female resident divided by the number of ways in which we can select any 2 residents.
A: There are different ways of solving this, I will do one that might help you to well understand what is going on.
You have a total of $39$ people. So the probability that you pick a male is $11/39$ and the probability that you pick a female is $28/39$.
Assume you pick one after the other. If you picked a male first then the probability of picking a female would be $28/38$ (since you already picked one person). Similarly, if you pick a female first, then the probability that you pick a male would be $11/38$. Since you pick them simultaneously the order doesn't matter and so you consider the two cases for which you get the pretended probability$$
P=\frac{11}{39}\times\frac{28}{38}+\frac{28}{39}\times\frac{11}{38}=2\times \frac{11}{39}\times\frac{28}{38}=\frac{308}{741}\simeq 0.416.
$$
A: Well based on strictly that you can draw a conditional probability tree diagram and add the possible outcomes (Which would be in our case male then female and female then male. 
Which will result in something looking like:
$$P(Male \cap Female)=P(male)P(female|male)+P(female)P(male|female)$$
$$\iff {11\over{39}}\times {28\over{38}}+{28\over{39}}\times{11\over{38}}={308\over741}$$
Hope this is helpful!
A: The word "simultaneously" has no relevance in this matter. 
Assume that they are chosen one by one and let $M$ denote the event that the first chosen is a male. Let $E$ denote the event that one of the chosen is male and one is female.
Then:
$$P(E)=P(E\mid M)P(M)+P(E\mid M^{\complement})=\frac{28}{38}\frac{11}{39}+\frac{11}{38}\frac{28}{39}$$
