A box contains 9 two-inch screws, of which 6 have a Phillips head and 3 have a r

Question

A box contains 9 two-inch screws, of which 6 have a Phillips head and 3 have a regular head. Suppose that you select 3 screws randomly from the box with replacement.
1) Find the probability there will be more than one Phillips head screw ?

According to a recent report, 60% of U.S. college graduates cannot find a full time job in their chosen profession. Assume 73% of the college graduates who cannot find a job are female and that 39% of the college graduates who can find a job are female.

2) Given a male college graduate, find the probability he can find a full time job in his chosen profession ?

Explanation / Answer

1.)

Total regular screws: 3

Probabaility of Philips Screw, P(P) = 6/9 = 2/3

Probabaility of Regular Screw, P(R) = 3/9 = 1/3

We are drawing 3 screws with replacement, So, Sample Space is:

[PPP, PPR, PRP, RPP, RRP, RPR, PRR, RRR]

Probabaility that there will more than 1 philips screw is:

P(P > 1) = P(PPP) + P(PPR) + P(PRP) +P(RPP)

Now, P(PPP) = (2/3)3 = 8/27

P(PPR) = (2/3)2 * 1/3= 4/27

P(PRP) = (2/3)2 * 1/3= 4/27

P(RPP) = (2/3)2 * 1/3= 4/27

P(P > 1) = 8/27 + (4/27)*3

P(P > 1) = 8/27 + 12/27

P(P > 1) = 20/27

This is desired probabaility

2)

P(not finding a Job) = 60% = 0.6

P(find a job) = 1 - 0.6 = 0.4

we need to find:

P(find a job | Male) = ?

Using Baye's theorem, we get

P(find a job | Male) = ( P(Male | find a job) * P(find a job) ) / P(Male)  

Let's make a cross tab:

Now, P(Male | find a job) = 0.244/0.4

P(find a job) = 0.4

P(Male) = 0.406

Putting in formula we get,

P(find a job | Male) =0.244/0.406 = 0.60098

This is the desired probability

not find job find job 0.6 0.4 women 0.438 0.156 (73 percent) (39 percent) men 0.162 0.244 (27 percent) (61 percent)

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