An individual is chosen from the maintenance staff of a company. Let X be the nu
ID: 3231158 • Letter: A
Question
An individual is chosen from the maintenance staff of a company. Let X be the number of years of service, and let Y be the hours of overtime worked per day. The following table gives the joint probability function of X and Y Draw up similar tables defining the conditional probability functions of Y given X = x, and X given Y = y. If a member of staff has been in service for 1 year, what is the expected number of hours of overtime worked per day. If a member of staff works 1 hour of overtime per day, what is the probability that he has been in service for at least 2 years?Explanation / Answer
Part-a
We are given the joint probabilities, so
the conditional probability P(Y/X=x)=P(Y=y,X=x)/P(X=x)
and the conditional probability P(X/Y=y)=P(Y=y,X=x)/P(Y=y)
x
Y: conditional probability P(Y/X=x)=P(Y=y,X=x)/P(X=x)
0.5
1.0
1.5
Total
1
(1/5)/(1/5+1/5+0)
=1/2
=0.5
(1/5)/(1/5+1/5+0)
=1/2
=0.5
(0)/(1/5+1/5+0)
=0
1.00
2
0/(0+3/10+1/10)
=0
(3/10)/(0+3/10+1/10)
=3/4
=0.75
(1/10)/(0+3/10+1/10)
=2/4
=0.25
1.00
3
0/(0+0+1/5)
=0
0/(0+0+1/5)
=0
(1/5)/(0+0+1/5)
=1
1.00
x
Y: conditional probability P(X/Y=y)=P(Y=y,X=x)/P(Y=y)
0.5
1.0
1.5
Total
1
(1/5)/(1/5+0+0)
=1
(1/5)/(1/5+3/10+0)
=2/5
=0.4
(0)/(0+1/10+1/5)
=0
2
(0)/(1/5+0+0)
=0
(3/10)/(1/5+3/10+0)
=3/5
=0.6
(1/10)/(0+1/10+1/5)
=1/3
3
(0)/(1/5+0+0)
=0
(0)/(1/5+3/10+0)
=0
(1/5)/(0+1/10+1/5)
=2/3
Total
1
1
1
Part-b
E(Y/X=1)=0.5*0.5 + 1.0*0.5 + 1.5*0 =0.75
Part-c
P(X>=2/Y=1)=0.6+0=0.6
x
Y: conditional probability P(Y/X=x)=P(Y=y,X=x)/P(X=x)
0.5
1.0
1.5
Total
1
(1/5)/(1/5+1/5+0)
=1/2
=0.5
(1/5)/(1/5+1/5+0)
=1/2
=0.5
(0)/(1/5+1/5+0)
=0
1.00
2
0/(0+3/10+1/10)
=0
(3/10)/(0+3/10+1/10)
=3/4
=0.75
(1/10)/(0+3/10+1/10)
=2/4
=0.25
1.00
3
0/(0+0+1/5)
=0
0/(0+0+1/5)
=0
(1/5)/(0+0+1/5)
=1
1.00
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