Answer to Question #304894 in Statistics and Probability for ced

Question

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).

Answer 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) = _nC_xp^x(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