Download ENGR 323 Beautiful Homework #4 Page 1 of 3 Leal Problem # 3

ENGR 323
Leal
Beautiful Homework #4
Problem # 3-16
Page 1 of 3
3-16. Some parts of California are particularly earthquake-prone. Suppose that in one such area,
20% of all homeowners are insured against earthquake damage. Four homeowners are to
be selected at random; let X denote the number among the four who have earthquake
insurance.
a. Find the probability distribution of X. [HINT: Let S denote a homeowner who has
insurance and F one who does not. Then 1 possible outcome is SFSS, with probability
(.2)(.8)(.2)(.2) and associated X value 3. There are 16 other outcomes.]
b. Draw the corresponding probability histogram.
c. What is the most likely value for X?
d. What is the probability that at least two of the four selected have earthquake
insurance?
Solution
In part (a) we’re asked to find the probability distribution, otherwise known as the
probability mass function (pmf), of the random variable X. We let X equal the number among
the four who are insured. The problem states that 20% of all homeowners are insured against
earthquake damage. It then tells us that from the sample space of all homeowners in California
we must choose four at random and find the probability of each of them being insured. For this
event we have two possible outcomes when we select homeowner, either that they are insured or
that they are not. We will let S equal a homeowner who is insured, and F equal one who is not.
We are told that there are 16 possible outcomes when we choose four homeowners at random.
We now can list all the possible outcomes for selecting four homeowners.
p(0) = P(X = 0) = P(FFFF) This is the probability that none of the homeowners is insured
p(1) = P(X = 1) = P(SFFF Υ FSFF Υ FFSF Υ FFFS) This is the probability that one homeowner
is insured. There are four possible ways
we can this probability, either the first
person chosen is insured and the rest are
not, or the second person chosen is insured
and the rest are not, etc. This probability
represents four disjoint events.
p(2) = P(X = 2) = P(SSFF Υ SFFS Υ SFSF Υ FSFS Υ FFSS Υ FSSF)
This is the probability that two are insured. This time we have six disjoint events that we can
have two homeowners insured.
p(3) = P(X = 3) = P(SSSF Υ FSSS Υ SFSS Υ SSFS) This is the probability that three are
insured.
p(4) = P(X = 4) = P(SSSS) This is the probability that all four selected are insured.
ENGR 323
Leal
Beautiful Homework #4
Problem # 3-16
Page 2 of 3
From the problem statement we are told P(F) = .8 and P(S) = .2, now we can determine the
probability distribution for the possible outcomes listed above.
p(0) = (.8)4 = .4096 For this first probability that zero homeowners are insured we are using
independence when we raise (.8) to the fourth power.
p(1) = 4[(.2)(.8)3] = .4096 For these next three probability we are using the fact that SFFF and
FSFF are mutually exclusive to add the probabilities.
p(2) = 6[(.2)2(.8)2] = .1536
p(3) = 4[(.2)3(.8)] = .0256
p(4) = (.2)4 = .0016
If we add up the probability for each of the five events they should equal 1.
We can display our results either as a list as we did above, in a table, or as a graph (See Figure
1.).
x
p(x)
0
.4096
1
.4096
2
.1536
3
.0256
4
.0016
p(x)
For part (b) we are asked to draw a probability histogram. This is the similar to the pmf, but
instead bars are used to display the pmf rather than lines. Figure 1. shows the resulting
histogram.
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
x = # of homeowners insured against earthquake damage
Figure 1. Probability histogram of insured homeowners
Part (c) asks for what values of p(x) is X largest, from our probability histogram above we can
see that p(x) is largest for x to 0 or 1.
ENGR 323
Leal
Beautiful Homework #4
Problem # 3-16
Page 3 of 3
Part (d) asks what is the probability that at least two of the four people select have earthquake
insurance. We can use the values from part a to determine the probability that at least two or
more people have earthquake insurance. We can write the answer by summing the probabilities
of x = 2,3, and 4.
P(X ≥ 2) = p(2) + p(3) +p(4) =.1808
We could calculate the answer by saying one minus the probability of X < 2:
1 − P(X<2) = 1 − p(0) + p(1) = .1808
Figure 2 shows the graph of the cumulative distribution function for the probabilities listed in
part (a). The cdf could be used ask questions about the probability of seeing values of x greater
than or less a certain given value of x.
1
0.8
F(x)
0.6
0.4
0.2
0
-1
0
1
2
3
4
x = # of homeowners insured against earthquake damage
Figure 2. cdf graph of insured homeowners
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