im getting confused in persentage question of probability..

a group consists of equal number of men and women of this group 20% of men and 50% of the women are unemployed. if a person is selected at random from this group the probability of the selected person being employed is ......... In this problem , lets take eg. 50 men and 50 women. Out of which 10 men are unemployed and 25 women unemployed..so probability of unemployed men is 1/10 and unemployed women is 1/2 .. but after that..i dont understand how to calculate the probability of selected person is employed . My ans is not coming correct

• Your selected person is either male or female, each with probability $\frac 12$. These cases are mutually exclusive. Hence the answer you want is $\frac 12 \times .8+\frac 12 \times .5$. – lulu Jun 30 '16 at 15:20

The answer is: P(guy|employed) = P(man).P(employed|male) + P(woman).P(employed|woman) = (0.5).(0.8) + (0.5).(0.5)

Another way of saying this:

Given that the person you picked is male, the probability of them being employed is $0.8$.

Given that the person you picked is female, the probability of them being employed is $0.5$.

The probability of the person you pick being male is $0.5$. Same thing for female.

So the probability is

$$P = P(\text{male})P(\text{employed, given male}) + P(\text{female})P(\text{employed, given female}) \\ = 0.5 \cdot 0.8 + 0.5 \cdot 0.5 = 0.65.$$