Probability Statistics

Question

Among the 12 applicants for managerial positions at a chain of movie theaters, 8 are college graduates and 4 are high school graduates. If three of the applicants are randomly selected,

a. Identify the possible outcomes in terms of whether a selected applicant is a college graduate (C) or a high school graduate (H).

b. Find the probability for each possible outcome and verify that these probabilities follow the two basic rules of the probabilities for the outcomes in a sample space.

Answer 1

Problem A

Possible Outcomes = {CCC, CCH, CHC, CHH, HCC, HCH, HHC, HHH}

Problem B

P(CCC)

= [C(8,3)C(4,0)]/C(12,3)

= [8!/[3!(8-3)!]]*[4!/[0!(4-0)!]]/[12!/[3!(12-3)!]]

= [(8*7*6)/(3*2*1)]/[(12*11*10)/(3*2*1)]

= (8*7*6)/(12*11*10)

= 56/220

≈ 0.255

P(CCH) <-- Note that this is the same as P(CHC) and P(HCC)

= [C(8,2)C(4,1)]/C(12,3)

= [[8!/[2!(8-2)!]]*[4!/[1!(4-1)!]]/[12!/[3!(12-3)!]]

= [(8*7)/(2*1)]*4/[(12*11*10)/(3*2*1)]

= (28*4)/(1320/6)

= 112/220

≈ 0.509

P(CHH) <-- Note that this is the same as P(HCH) and P(HHC)

= C(8,1)C(4,2)/C(12,3)

= 8[4!/2!(4-2)!]/220

= 8[(4*3)/(2*1)]/220

= (8*6)/220

= 48/220

≈ 0.218

P(HHH)

= C(8,0)C(4,3)/C(12,3)

= [4!/[3!(4-3)!]]/220

= 4/220

≈ 0.018

We can see that 56/220 + 112/220 + 48/220 + 4/220 = 220/220 = 1, so it verifies the basic rules of the probabilities for the outcomes in a sample space.

Hope this helped!

Answer 2

possible outcomes are

3 college grads 0 HS which can happen only 1 way: ccc

2 college grads 1 HS which can happen 3 ways: cch, chc, hcc

1 college grad 2 HS which can happen 3 ways: chh, hch, hhc

0 college grads 3 HS which can happen only 1 way: hhh

4 possible outcomes or combinations, but 1+3+3+1 permutations if order matters

P(3c0h) = 8/12 x 7/11 x 6/10 = 2/3 x 7/11 x 3/5 = 42/165= 14/55

P(2c1h) = 8/12 x 7/11 x 4/10 = 2/3 x 7/11 x 2/5 = 28/165

+ 8/12 x 4/11 x 7/10 = 28/165

+ 4/12 x 8/11 x 7/10 = 28/165

=3 x 28/165 = 84/165 = 28/55

P(1c2h) = 8/12 x 4/11 x 3/10 = 2/3 x 4/11 x 3/10 = 4/55

+ 4/12 x 3/11 x 8/10 = 2/3 x 3/11 x 4/5 = 4/55

+ 4/12 x 8/11 x 3/10 = 2/3 x 8/11 x 3/10 = 4/55

= 12/55

P(3h) = 4/12 x 3/11 x 2/10 = 2/3 x 3/11 x 1/5 = 1/55

(14+28+12+1)/55 = 55/55= 1

sum of the probabilities of each outcome = 1 = 100%