Solution ST361 Quiz 2, 2/10/2009
1. (40pts) An engineer is examining three products in a row to see if they will
meet design specifications. Do the following:
(a) Find the sample space of this “experiment” (you may use “M” if a
product meets the specifications and use “F” if it fails to meet the
specifications).
S={MMM, MMF, MFM, FMM, FFM, FMF, MFF, FFF}
(b) Define A to be the event that the second product meets the specifications.
Express A as a subset of the sample space.
A={MMM, FMF, MMF, FMM}
(c) Define B to be the event that the third product meets the specifications.
Express B as a subset of the sample space.
B={MMM, FFM, FMM, MFM}
(d) Define C to be the event that at least two products meet the
specifications. Express C as a subset of the sample space.
C={MMM, MFM, MMF, FMM}
(e) Find A  B, A  B, A  C , B  C , A  B  C.
A  B  {MMM, FMM}
A  B ={MMM, FMM, FMF, MMF, FFM, MFM}
A  C = {MMM, FMM, MMF}
B  C ={MMM, FMM, MFM}
A  B  C ={MMM, FMM}
2. (30pts) There are 2 white balls, 3 yellow balls and 1 blue ball in an urn. Two
balls are randomly selected from the urn one after the other. Define A to be
the event that the first selected ball is yellow and B to be the event that the
second selected ball is white. Do the following:
(a) Find the probability that the first selected ball is yellow, that is, find
P(A).
Since there are 3 yellow balls out of total 6 balls, so P(A)=3/6=1/2.
(b) Given the first selected ball is yellow, find the probability that the
second selected ball is white, that is, find P(B|A).
Given A has occurred, there are only 5 balls left, out of which there
are 2 white balls. So P(B|A)=2/5
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(c) Find the probability that first selected ball is yellow and the second
selected ball is white, that is, find P( A  B)
P( A  B) = P(A)P(B|A) (by the multiplicative law) = (1/2)*(2/5) = 1/5 =
0.2.
3. (30 pts) A discrete random variable x may take values 0, 1, 2, 3, and has
the probability mass function given by the following table:
x
P(x)
0
1
2
3
0.2
0.1*(k+1)
0.3*(k-1)
0.2
Do the following:
(a) Find value k such that the above table gives a probability mass
function.
 P( x)  1  0.2  0.1(k  1)  0.3(k 1)  0.2  1 . So 0.4k = 0.8; k=2.
x
(b) Find P[ x  2] and P[x>1].
From (a), we have P(1) = 0.3, P(2) = 0.3.
P[ x  2] = P[x=0] + P[x=1] + P[x=2] = 0.2 + 0.3 + 0.3 = 0.8
P[X>1] = P[x=2] + P[x=3] = 0.3 + 0.2 = 0.5
(c) Find the expected value of x.
 x =0*0.2 + 1 * 0.3 + 2 * 0.3 + 3 * 0.2 = 1.5
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