A company has 7 applicants, three women and four men. Suppose that the 7 applicants are equally qualified and that no preference is given for choosing either gender. Let X be the random variable described by the number of women chosen to fill the two positions. (a) Express the probability distribution of X as a formula in combinatorial notation. (b)Find P(X = 1). (c) Find P(X ≤ 1).

(a)

There are three women out of the seven applicants, therefore, the probability of selecting a woman p = 3/7 (probability of success)

Also, there are only two possible outcomes in a single trial: either selecting a woman or a man, the we say that the random variable X follows a binomial distribution given by:

P(X=x) = nCxpx(1-p)n-x

where

n = number of trials

x = number of times for a specific outcome within n trials

p = probability of success on a single trial

C = combination

(b)

P(X=1)

Here, we have x=1 and n=2

P(X=1) = 2C1(3/7)1(1-3/7)

= 0.4898

(c)

P(X≤ 1)

Here, x=0 or 1 and n=2

P(X≤ 1) = P(X=0) + P(X=1)

=2C0(3/7)0(1-3/7)2 + 2C1(3/7)1(1-3/7)

= 0.3265 + 0.4898

= 0.8163