Question

(20) 5. A company has 7 applicants for 3 positions. Two are women and five are men. Suppose the applicants are equally qualified, and no preference is given for choosing either gender. Let x equal the number of women chosen to fill the 3 positions. Show your work. (4) a. Find the probability that both women and one man are hired. (6) b. Construct a bar chart for w = number of women chosen. (Hint: w=0, w = 1, w = 2). (6) c. Construct a bar chart for m = number of men chosen. (Hint m=0, m = 1, m = 2, or m = 3). (2) d. Find the probability that all three men are hired and the probability that no men are hired.

Answer 1

Total number of applicants = 7

Number of men = 5

Number of women = 2

Number of positions = 3

So, number of selections that can be made to fill the 3 positions from the 7 applicants = sample size = n (S) = ⁷C₃ = 7! / (3! * 4!) = 35

Answer (a) :

Let E be an event of selecting two women and one man.

So, n (E) = ²C₂ * ⁵C1 = 1 * 5 = 5

Therefore, P (E) = n(E) / n(S) = 5 / 35 = 0.1428

Answer (b):

We use similar method as used in the above question to get the answer of this question.

P (no women is hired ) =(  ²C₀ * ⁵C₃ ) / n(S) = (1 * 10) =  10 / 35 = 0.2857

P ( one women is hired ) =(  ²C₁ * ⁵C₂ ) / n(S) = (2 * 10) = 20 / 35 = 0.5714

P ( two women are hired ) =(  ²C₂* ⁵C₁ ) / n(S) = (1 * 5 ) = 5 / 35 = 0.1428

W = number of women hired	0	1	2
P ( W = w )	10 / 35	20 / 35	5 / 35

P (W = w) 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0 1 2

Answer (c) :

P (Choosing no men ) = 0 ; This is because there are three positions and only 2 women candidates have applied. So, Atleast 1 man has to be hired.

P ( Choosing one men ) = ( ⁵C₁ * ²C₂ ) / n(S) = ( 5 * 1 ) / 35 = 5 / 35 = 0.1429

P ( Choosing two men ) = ( ⁵C₂ * ²C₁ ) / n(S) = ( 10 * 2 ) / 35 = 20 / 35 = 0.5714

P ( Choosing three men ) = ( ⁵C₃ * ²C₀ ) / n(S) = ( 10 * 1 ) / 35 = 10 / 35 = 0.2857

M = Number of men	0	1	2	3
P ( M = m)	0	0.1429	0.5714	0.2857

P(M= m) 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3

Answer (d):

Let E1 and E2 be two events of hiring three men and no men, respectively.

So, n (E1) = ⁵C₃ * ²C₀= 10 * 1 = 10

and, n (E2) = 0 ; since, here are three positions and only 2 women candidates have applied. So, Atleast 1 man has to be hired.

Therefore, P ( E1 ) = n(E1) / n(S) = 10 / 35 = 0.2857

and, P (E2) = n(E2) / n(S) = 0 / 35 = 0