KEY - BAN530 Discrete (BINOMIAL AND POISSON) Probability Handout
BINOMIAL EXAMPLES
BAN 530 (page 3 of handout)
1.
If X ~ b(8, .7), find the following probabilities:
(a)
P(X = 2)
P(X = 2) = P(X ≤ 2) - P(X ≤ 1) = 0.0113 – 0.0013 = 0.01


2.
3.
(b)
P(X 
(c)
P(X > 7) = 1 – P(X ≤ 7) = 1 – 0.9424 = 0.0576
In a certain city, 40% of the residents are opposed to the widening of Main Street. In a random
sample of 20 residents, what is the probability that the number opposed to the widening is
If X ~ b(20, .4),
(a)
more than ten
P(X > 10) = 1 – P(X ≤ 10) = 1 – 0.8725 = 0.1275
(b)
between 15 and 18, inclusive
P(15 ≤ X ≤ 18) = P(X ≤ 18) – P(X ≤ 14) = 1 (rounded) – 0.9984 = 0.0016
(c)
fewer than eight
P(X < 8) = P(X ≤ 7) = 0.4159
(d)
no more than 13?
P(X ≤ 13) = 0.9935
Suppose that 70% of a certain population of drivers regularly use seat belts. You take a random
sample of 15 of these drivers. What is the probability that the number regularly using seat belts is
If X ~ b(15, .7),
(a)
more than 14
P(X > 14) = 1 - P(X ≤ 14) = 1 - 0.9953 = 0.004
(b)
exactly 10?
P(X = 10) = P(X ≤ 10) - P(X ≤ 9) = 0.4845 – 0.2784 = 0.2061
BINOMIAL PROBABILITY DISTRIBUTION PROBLEMS (page 4 of handout)
1.
Suppose thirty percent of the voters in a large voting district are veterans. (a) If twenty voters are
randomly selected, find the probability that less than five will be veterans. (b) Out of a sample of
twenty, how many would you expect to be veterans?
(a) X ~ b(20, 0.30) P(X < 5) = P(X ≤4) = 0.2375
(b) E(X) =  = n 
2.
Suppose sixty percent of the voters in a large town are opposed to the construction of a new road. If
twelve voters are randomly selected, what is the probability that exactly seven of them are opposed
to the construction of the new road?
X ~ b(12, 0.60)
P(X = 7) = P(X ≤ 7) - P(X ≤ 6) = 0.5618 – 0.3348 = 0.227
3.
Suppose forty percent of the student body at a SHSU are in favor of allowing the consumption of
alcohol in the dormitories (for those 21 and older). If ten students are randomly selected, what is the
probability that more than seven of them are opposed to the consumption of alcohol in the dorms?
X ~ b(10, 0.60)
P(X > 7) = 1 - P(X ≤ 7) = 1 - 0.8327 = 0.1673
4.
Suppose 40% of all people have type A blood. If a random sample of twelve people, what is the
probability that five have type A blood?
X ~ b(12, 0.40)
P(X = 5) = P(X ≤ 5) - P(X ≤ 4) = 0.6652 – 0.4382 = 0.227
POISSON PROBABILITY DISTRIBUTION PROBLEMS (page 5 of handout)
1.
Suppose a hospital averages three patient accidents per month.
(a)
What is the probability that four patients will have accidents in the next month?
E(X) =  = expected number of accidents in the next month = 3
X = # of accidents in the next month
X ~ P(3)
P(X = 4) = P(X ≤ 4) - P(X ≤ 3) = 0.8153 – 0.6472 = 0.1681
(b)
What is the probability that more than four accidents occur in the next month?
E(X) =  = expected number of accidents in the next month = 3
X = # of accidents in the next month
X ~ P(3)
P(X > 4) = 1 - P(X ≤ 4) = 1 – 0.8153 = 0.1847
(c)
What is the probability that there are at most six accidents in the next two months?
E(X) =  = expected number of accidents in the next two months = 6
X = # of accidents in the next two months
X ~ P(6)
P(X ≤ 6) = 0.6063
BAN530
POISSON PROBABILITY DISTRIBUTION EXAMPLES (page 6 of handout)
1.
Suppose, according to historical evidence, the number of major earthquakes occurring in
California averages five per year.
(a)
What is the probability that California will have seven major earthquakes in one year?
E(X) =  = expected number of earthquakes in one year period = 5
X = # of earthquakes in a one year period
X ~ P(5)
P(X = 7) = P(X ≤ 7) - P(X ≤ 6) = 0.8666 – 0.7622 = 0.1044
(b)
2.
What is the probability that California will have exactly five major earthquakes in a two
year period? (YOUR COPIES OF THE POISSON TABLE DO NOT INCLUDE THIS
PAGE— SORRY I WILL BRING TO CLASS NEXT WEEK)
E(X) =  = expected number of earthquakes in two year period = 10
X = # of earthquakes in a two year period
X ~ P(10)
P(X = 5) = P(X ≤ 5) - P(X ≤ 4) = 0.0671 – 0.0293 = 0.0378
Suppose that in New York City, on the average, 3.5 empty taxicabs pass a busy street corner every
15 minutes.
(a)
What is the probability that no empty taxicabs pass in the next 15 minutes?
E(X) =  = expected number of empty taxicabs that pass a busy street corner every 15 minutes = 3.5
X = # of empty taxicabs that pass a busy street corner in 15 minutes
X ~ P(3.5)
P(X= 0) = 0.0302
(b)
What is the probability that 5 or more empty taxicabs pass in the next 30 minute interval?
E(X) =  = expected number of empty taxicabs that pass a busy street corner every 30 minutes = 7
X = # of empty taxicabs that pass a busy street corner in 30 minutes
X ~ P(7)
P(X ≥ 5) = 1 – P(X ≤ 4) = 1 – 0.1730 = 0.827
3.
Americans receive an extraordinary amount of junk mail every day. Suppose you estimate that the
mean amount of junk mail you receive daily is two pieces.
(a)
What is the probability that you will not receive any junk mail three days in a row?
E(X) =  = expected number of pieces of junk mail you receive in 3 days = 6
X = # of pieces of junk mail you receive in 3 days
X ~ P(6)
P(X = 0) = 0.0025
(b)
What is the probability that you receive four pieces of junk mail in a two day period?
E(X) =  = expected number of pieces of junk mail you receive in 2 days = 4
X = # of pieces of junk mail you receive in 2 days
X ~ P(4)
P(X = 4) = P(X ≤ 4) - P(X ≤ 3) = 0.6288 – 0.4335 = 0.1953
BAN 530 (page 7 of handout)
Discrete Probability Examples
1.
Suppose twenty percent of the registered voters in a large metropolitan city support the reelection
of the mayor. If a random sample of twenty of the city’s registered votes is selected, what is the
probability that fewer than four support the reelection of the mayor?
Success = support the reelection of the mayor
 = P(support the reelection of the mayor) = 0.20
n = 20 registered voters
X = # who support the reelection of the mayor out of the 20 voters
X ~ b(20, 0.20)
P(X < 4) = P(X ≤ 3) = 0.4114
2.
A local car wash boasts an average of nine cars per hour arrive at the car wash on Saturday
afternoons. What is the probability, in a one hour period on a Saturday afternoon, that at least six
cars arrive at the car wash?
E(X) =  = expected number of arrivals in a one hour period = 9
X = # of arrivals in a one hour period
X ~ P(9)
P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – 0.1157 = 0.8843
3.
Suppose forty percent of Americans eat at a fast-food restaurant on a regular basis. In a random
sample of 15 Americans, what is the probability that seven regularly eat at a fast-food restaurant?
Success = regularly eat at a fast-food restaurant
 = P(regularly eat at a fast-food restaurant) = 0.40
n = 15 Americans
X = # who regularly eat at a fast-food restaurant out of the 15
X ~ b(15, 0.40)
P(X = 7) = P(X ≤ 7) - P(X ≤ 6) = 0.7869 – 0.6098 = 0.1771
4.
Suppose it is known that thirty percent of MBA candidates have an undergraduate degree in a
discipline outside of business. In a random sample of ten MBA candidates, what is the probability
that more than seven have an undergraduate degree in business?
Success = have an undergraduate degree in business
 = P(have an undergraduate degree in business) = 0.70
n = 10 MBA candidates
X = # who have an undergraduate degree in business out of 10
X ~ b(10, 0.70)
P(X > 7) = 1 - P(X ≤ 7) = 1 - 0.6172 = 0.3828
5.
Suppose a manufacturing plant knows that the typical number of defects in a lot of 10,000 is 4.2.
What is the probability that in the next two lots of 10,000, six or fewer are defects?
E(X) =  = expected number of defects in the next two lots of 10,000 = 8.4
X = # of defects in the next two lots of 10,000
X ~ P(8.4)
P(X ≤ 6) = 0.2670
6.
A bank reveals the mean number of ATM transactions in a fifteen minute period is 3.2 for a new
ATM machine on the campus of a local university. In the next thirty minutes, what is the
probability there are five to nine transactions at the new ATM machine?
E(X) =  = expected number of ATM transactions in a thirty minute period = 6.4
X = # of ATM transactions in a thirty minute period
X ~ P(6.4)
P(5 ≤ X ≤ 9) = P(X ≤ 9) – P(X ≤ 4) = 0.8858 – 0.2351 = 0.6507
7.
Suppose four out of five dentists recommend a particular sugarless gum for their patients. If a
random sample of twelve dentists is taken, what is the probability that eight recommend the gum
for their patients?
Success = recommend a particular sugarless gum for their patients
 = P(recommend a particular sugarless gum for their patients) = 0.80 (4 out of 5)
n = 12 dentists
X = # who recommend a particular sugarless gum for their patients out of 12
X ~ b(12, 0.80)
P(X = 8) = P(X ≤ 8) - P(X ≤ 7) = 0.2054 – 0.0726 = 0.1328
8.
Suppose a restaurant claims that, on the average, sixteen people arrive to be seated in a twenty
minute interval on a Saturday evening. In a five minute interval on a Saturday evening, what is
the probability that more than four people arrive to be seated?
E(X) =  = expected number of arrivals in a five minute interval = 4
X = # of arrivals in a five minute interval
X ~ P(4)
P(X > 4) = 1 – P(X ≤ 4) = 1 – 0.6288 = 0.3712
9.
Suppose it is known that forty percent of the student population at a large public university is
female. Out of a random sample of twenty-five students, what is the probability that at least nine
are male?
Success = male
 = P(male) = 0.60
n = 25 students
X = # who are male out of the 25 students
X ~ b(25, 0.60)
P(X ≥ 9) = 1 - P(X ≤ 8) = 1 - 0.0043 = 0.9957
10. Suppose, on average, there are two potholes per mile on a county road. What is the probability
there are less than four potholes in a two-mile section of the county road?
E(X) =  = expected number of potholes in a two-mile section of the county road = 4
X = # of potholes in a two-mile section of the county road
X ~ P(4)
P(X < 4) = P(X ≤ 3) = 0.4335