CS 3341 HW #4 Solution
1. Suppose that the probability of a customer buying a hamburger at Burger Barn is 0.60. If
there are 15 customers in line and 10 hamburgers already prepared, what is the probability
that a customer will have to wait for a hamburger?
Let X = # customers in the line who want to buy a burger. Then X has Binomial (n=15,
p=10) distribution.
15 15 
So, P(A customer has to wait) = P(X ≥ 11) = ∑  (0.60 )k (1 − 0.60 )15−k = 0.217
k =11  k 
2. Suppose that the probability of a female birth is 0.5. A couple desires to have a girl. They
have decided to keep having children until they have exactly one girl. What is the probability
that the family will have exactly three children?
Let X = # kids born in the family till the first girl child is born (including the girl). Then X
has Geometric (p = 0.5) distribution.
So, P(Family has exact three children) = P(X = 3) = (1-0.5)2 (0.5) = 0.125
3. A company produces and ships 16 PC’s knowing that four of them have defective wiring.
The company that has purchased the computers is going to thoroughly test three of the
randomly selected computers. The purchasing company can detect the faulty wiring. What is
the probability that the purchasing company will find three defective computers?
Let X = # defective PC’s in the sample. Think of a defective as a success. Then X has a
Hypergeometric distribution with population size N = 16, # successes in the population M =
4, and sample size n = 3.
 4 12 
  
3 0
So, P(X = 3) =    = 0.007
16 
 
3
4. A manufacturing company produces 10,000 plastic mugs per week. This company supplies
mugs to another company, which packages the mugs as part of picnic sets. The second
company randomly samples 10 mugs sent from the supplier. If two or fewer of the sampled
mugs are defective, the second company accepts the lot. What is the probability that the lot
will be accepted if the mug manufacturing company actually is producing 10% defective
mugs?
Let X = # defective mugs in the sample. This exercise is similar to the previous one, but here
the population size N = 10,000 is large compared to the sample size n = 10. So, we can use
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the Binomial approximation to Hypergeometric. Thus, here X follows Binomial (n=10,
p=0.10) distribution.
Now, P(Lot is accepted) = P(X ≤ 2) =
2 10 
∑ 

k =0 
k
(0.10)k (1 − 0.10)10−k
= 0.93
5. In the United States, raising money for charitable causes is a competitive activity. Some
causes receive much more support than others. For example, almost 53% of all U.S.
households contribute to some religion. Also, 24% of all households give to some health
cause. Education receives charitable contributions from about 15% of all U.S. households.
Suppose a professional fund raiser conducts a nationwide telephone survey of 12 randomly
selected U.S. households. What is the probability that exactly 10 of these households
contributed some to health cause?
Let X = # households in the sample that contributed to some health cause. Then X follows a
Binomial (n = 12, p = 0.24) distribution. (This exercise gives you much more information
than what you really need.)
12 
Now, P(X = 10) =  (0.24 )10 (1 − 0.24 )12−10 = 0.000024
10 
6. A city has 18 police officers who are eligible for promotion. Eleven of the 18 are from race
A. Suppose five police officers are chosen for promotion, and only one officer among them is
from race A. If the officers chosen for promotion had been randomly selected, what is the
probability that one or fewer of the five promoted officers would have been from race A?
What might this result indicate?
Let X = # race A officers in the sample of 5 who are chosen for promotion. Think of being
from race A as a success. Then X has a Hypergeometric distribution with population size N =
18, # successes in the population M = 11, and sample size n = 5.
11 7  11 7 
   +   
0 5
1 4
P(X ≤ 1) =       = 0.045
18 
 
5
Since, this probability is low, there is some evidence of discrimination in the promotion
against race A officers.
7. A medical researcher estimates that 0.00004 of the population has a rare blood disorder. If
the researcher randomly selects 100,000 people from the population, what is the probability
that more than 10 people will have this disorder? Suppose that the researcher does get more
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than 10 people with this disorder in the sample of 100,000 people, but the sample was taken
from a particular geographic region. What might the researcher conclude from the results?
(Make appropriate assumptions.)
Let X = # people in the sample of n=100,000 people who suffer from the disease. This is a
Binomial random variable with large and small p (= 0.00004). We can approximate this
distribution using a Poisson (λ = np = 4). So,
P(X > 10) = 1 – P(X ≤ 10) = 1 –
10
∑
k =0
e −4 4 k
= 1 – 0.997 = 0.003
k!
Thus, we see that the probability of finding more than 10 patients is very small when there
prevalence of the disease is p = 0.00004. However, if the researcher does get more than 10
people in a geographical area with this disease, it would indicate the prevalence of the
disease in that area may be higher than 0.00004.
8. On Monday mornings, a bank has only one teller window open for deposits and withdrawals.
Experience shows that the number of customer arrivals has a Poisson distribution and the
customers arrive at the rate of 2.8 customers in a 4-minute interval. The teller can serve no
more than 5 customers in any 4-minute interval at this window. What is the probability that,
during any given 4-minute interval, the teller will be unable to meet the demand? When the
demand cannot be met during any given 4-minute interval, a new window is opened. What
percentage of time will a second window have to be opened?
Let X = # customers arriving at the teller in the given 4-minute interval. It is given that X
follows Poisson (λ = 2.8) distribution.
P(Teller is unable to meet the demand) = P( X > 5) = 1 – P(X ≤ 5) = 1-
5
∑
k =0
e −2.8 2.8 k
k!
= 1 – 0.935 = 0.065
P(Second window open) = P( X > 5) = 0.065.
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