In: Statistics and Probability
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 ?
1.)
Total screws: 9 Total Philips screws: 6
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:
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) |
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