Question: A company has five applicants for two positions: two women and three men. Suppose that the five applicants are equally qualified and that no preference is given for choosing either gender. Let x equal the number of women chosen to fill the two positions.
a. Write the formula for p( x), the probability distribution of x.
b. What are the mean and variance of this distribution?
c. Construct a probability histogram for x.  Edit

Answer: a. Probability distribution of x

P(x=0)[prob. of no women]=3c2/5c2=3/10
P(x=1)[prob. of one women]=(2c1*3c1)/5c2=6/10
P(x=2)[prob. of 2 women]=2c2/5c2=1/10

x: 0 1 2
P(x): 3/10 6/10 1/10

b. mean=E(x)=0*3/10+1*6/10+2*1/10=6/10+2/10=8/10
E(x^2)=0^2*3/10+1^2*6/10+2^2*1/10=0+6/10+4/10=1
variance=E(x^2)-{E(x)}^2=1-(8/10)^2=1-64/100=36/100=6/10=3/5
  Edit