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L. Discrete Distributions.
1. Binomial Distribution.
L1. Text 5.18-5.21, 5.23, 5.27[5.14a-d, 5.15, 5.16, 5.19, 5.20, 5.21] (5.15, 5.16, 5.18*, 5.19, 5.20).
2. Geometric Distribution.
L2-L5.
3. Poisson Distribution.
L6-L8. Text 5.30, 5.31, 5.33, 5.34, 5.35, 5.38[5.29 – 5.30, 5.34 - 5.37] (5.22 - 5.23, 5.26-5.29).
4. Hypergeometric Distribution.
Text 5.44-5.45, 5.49-5.50 [5.22 – 5.23, 5.27 – 5.28] (5.30, 5.31, 5.35, 5.36). L9, L10.
5. Review.
Note that problem L11 is a review of the continuous uniform distribution.
This document includes Binomial and Geometric distribution problems.
Binomial Distribution Problems
Exercise 5.18 [Exercise 5.14a-c in 8th and 9th]: Do the following .
a) Px  0 when n  4 and p  .12 . Solution: If we use the formula Px  C xn p x q n x and recall that
C xn 
n!
and q  1  p , then P0  C 04 p 0 q 4  11.88 4  .5997 .
n  x !x!
b) Px  9 when n  10 and p  .40 . Solution: If we use the binomial table, P9  Px  9  Px  8
 .99990  .99832  .0016 .
c) Px  8 when n  10 and p  .50 . Solution: If we use the binomial table, P8  Px  8  Px  7
 .98926  .94531  .04395
d) Px  5 when n  6 and p  .83 . Solution: If we use the formula Px  C xn p x q n x , P5  C56 p 5 q1

6!
.83 5 .17 1  6.39390 .17   .4018
5! 1!
e) (in 9th edition) Px  9 when n  10 and p  .90 . Solution: If we use the binomial table,
P9  Px  9  Px  8 . But we are in trouble because we do not have a table for p  .90 . But 9
successes out of 10 tries means one failure out of 10 tries and the probability of failure is p  .10 .
P1  Px  1  Px  0  .73610  .34868  .3874 . Since I don’t have an answer book for this problem, I
will also do P9  C910 p 9 q1 
10!
.99 .11  10 .38742 .1  .3874
9! 1!
Exercise 5.19 [Not in 8th and 9th]: If n  5 and p  .4 find the following.
a) Px  4 . Solution: P4  Px  4  Px  3  .98976  .91296  .0768
a) Px  3 . Solution: Px  3  .9130
a) Px  2 . Solution: Px  2  Px  1  .3370
a) Px  1 . Solution: Px  1  1  Px  1  1  .33696  .6630
Problem L1: Assume that x has the binomial distribution with 10 tries and that the probability of
success on any one try is 25%. Find the following:
a. P4
b. E x  , Var x 
c. The probability that x is at least 1, Px  1 .
d. If the probability of success on any one try is .75, and n  10 , what is P6 ?.
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e. Using the cumulative binomial table find the following:
(i) Px  5 when n  10 , p  .2
(ii) Px  5 when n  10 , p  .2
(iii) P2  x  8 when n  10 , p  .4
(iv) P3  x  9 when n  10 , p  .7
f. If ten people are selected at random from a large (continuous) population
what is the probability that at least one makes more than the median income?
g. In f., what is the probability that between two and eight make more than the median
income?
h. What is the probability that all make more than the median income.?
i. For a binomial distribution with n  10 , p  .4 , find the smallest symmetrical region
about the mean that contains (at least) 95% of the probability.
Solution: If p  .25 and n  10 , we can write x ~ B.25,10 
a) If we use the formula Px  C xn p x q n x and recall that C xn 
P4  C 410 p 4 q 6 
n!
and q  1  p , then
n  x !x!
10!
.25 4 .75 6  10  9  8  7 .00391 .17798   .1460 . You could also use the table:
6! 4!
4  3  2 1
P4  Px  4  Px  3  .92187  .77588  .14599 .
b) The formula for the mean is Ex   np , Varx   npq (where q  1  p ; in this case 1  .25  .75 ), so
  .2510  =2.5,  2  npq  2.5.75  1.875 .
c) The probability that x is at least 1, Px  1 is the probability that x is any number equal to or above 1
and is thus the sum of P1  P2  P3    P10  . But, remember, that 'at least one' is the
complement of 'none', so Px  1  1  P0  1  C 010 .25 0 .75 10  1  .05631  .94369 .
10!
.75 6 .25 4  .1460 .This is the same as the answer for a)
4! 6!
and could be gotten from the table the same way. To see why, read on!
e) (i) Px  5 when n  10 , p  .2 x ~ B.2,10  . Px  5  Px  5  Px  4
 .99363  .96721  .02642 .
(ii) Px  5 when n  10 , p  .2 . According to the table, this is .99363.
(iii) P2  x  8 when n  10 , p  .4 The outline says that for any discrete distribution (with integer
values for x ) Px1  x  x 2   F x 2   F x1  1  Px  x 2   Px  x1  1 . So P2  x  8
 Px  8  Px  1  .99832  .04636  .95196 .
(iv) P3  x  9 when n  10 , p  .7 . You cannot find this on the table, because you can use the
probability of failure, .3, instead of .7. In 10 tries, 3 successes correspond to 7 failures and 9 successes,
correspond to 1 failure. So find P1  x  7  when n  10 , p  .3 . P1  x  7   Px  7  Px  0
 .99841  .02825  .97016 .
f) If people are selected from a relatively large population, the probability that each individual has an
income above the median is .5. Since there are 10 people, x ~ B.5,10  . Px  1  1  P0
 1  .00098  .99902
g) x ~ B.5,10  again. P2  x  8  Px  8  Px  2  .98926  .01074  97852 .
h) x ~ B.5,10  again. P10   Px  10   Px  9  1  .99902  .00098 .
i) For x ~ B.4,10  we are supposed to find the smallest symmetrical region about the mean that contains (at
least) 95% of the probability. The particular intervals that we must include, should have equal probabilities
above and below them. Since this means that 5% must be split between these two areas, we want x1 so that
Px  x1   .025 and Px  x 2   .025 For 1, Px  1  Px  0  .00605 and
d) p  .75 and n  10 . P6  C610 p 6 q 4 
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Px  2  Px  1  .04636 . On the high side, if we try Px  9  1  Px  9  1  .99990  .00010 ,
Px  8  1  Px  8  1  .99832  .00168 , Px  7  1  Px  7  1  .98771  .01229 ,
Px  6  1  Px  6  1  .94524  .05476 . So the lowest number we can use is 1 and the highest number
we can use is 7. The interval we have found is hardly symmetrical, but it is the best we can do.
Exercise 5.20 [Exercise 5.15 in 8th and 9th]: Find the mean and standard deviation if x has the binomial
distribution and a) n  4 and p  .10 , b) n  4 and p  .40 , c) n  5 and p  .80 and d) n  3 and
p  .50.
Solution: According to the outline, for the binomial distribution Px  C xn p x q n x , where p is between
zero and one and is the probability of success on an individual try, q  1  p is the probability of failure and
n is the number of tries. The formula for the mean is Ex   np .The variance is Varx   npq .
a) n  4 and p  .10. So   4.10   0.40 , q  1  .10  .90 ,  2  npq  .40.90  0.36 and
  0.36  0.60 .
b) n  4 and p  .40. So   4.40   1.60 , q  1  .40  .60 ,  2  npq  1.60.60  0.96 and
  0.96  0.9798 .
c) n  5 and p  .80. So   5.80   4.00 , q  1  .80  .20 ,  2  npq  4.00.20  0.80 and
  0.80  0.8944.
d) n  3 and p  .50. So   3.50   1.50 , q  1  .50  .50 ,  2  npq  1.50.50  0.75 and
  0.75  0.8660
Exercise 5.21 [Exercise 5.16 in 8th and 9th]: Assume that there is a 50% chance of a stock yielding an
increased closing price on any given day. What is the probability of an increase in 5 consecutive days?
Solution: This answer is copied from the Instructor’s Solutions Manual and is unchecked. Given p = 0.5
and n = 5, P(X = 5) = 0.0312. You should be able to get this by looking up Px  5 and Px  4 on the
binomial table and subtracting them.
(Exercise 5.18 In 8th edition only): Suppose that a random sample of web surfers is selected and ask if
they remember a specific internet banner. Use the binomial model and assume that p  0.4.
a. What is the expected value or mean of this binomial distribution?
b. What is the standard deviation of this binomial distribution?
c. What is the probability that none of the surfers will remember the banner ad?
d. What is the probability that exactly one of the surfers will remember the banner ad?
e. What is the probability that exactly two of the surfers will remember the banner ad?
f. What is the probability that three or more of the surfers will remember the banner ad?
Solution: n  5 and p  .4. q  1  p  .6. Px  C xn p x q n x
a)   np  5.40   2.00.
b)  2  npq  2.00.60  1.20 .   1.20  1.0954.
c) P0   C 05 p 0 q 5 
5!
.40 .65  .07776 . Better, if we use Table 15, the Cumulative Binomial
5!0!
Distribution, F 0  Px  0  .07776 .
d ) P1  C15 p 1 q 4 
5!
.41 .64  5 .4.1296   .25920 . Better, if we use the Cumulative
4!1!
1
Binomial Table , Px  1  Px  1  Px  0  .33696  .07776  .2590 .
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5!
.42 .63  5  4 .16 .216   .34560 . Better, if we use the Cumulative
3!2!
2 1
Binomial Table , Px  2  Px  2  Px  1  .68256  .33696  .34560 .
e) P2  C 25 p 2 q 3 
f) To find Px  3 , compute P3  P4 . Better, using the table, find
Px  3  1  Px  2  1  .68256  .31744 .
[Exercise 5.19 in 8th and 9th edition only] A telephone company has a 70% chance of making a same day
repair. For the first five repair requests find: a) the probability that all five repairs will be made on the same
day; b) the probability that at least 3 repairs will be made on the same day; c) the probability that fewer
than two repairs will be made on the same day. d) In a) – c) what assumptions are required? e) what are the
mean and standard deviation of the distribution? f) Repeat a) – c) and e) with an 80% probability and e)
compare the results with a) – c) and e).
Solution: This is a binomial problem with a sample size of 5 and a probability of success of .7.
5!
.7 5 .30  1.16807 1  .16807 . To
a) All five repaired on the same day: P5  C 55 p 5 q 0 
0!5!
do this using the table, note that we cannot use p  .7 , because the table only goes up to 5. But, if
we want 5 successes out of 5 tries, we want 0 failures. The probability of failure is
q  1  p  1  .7  .3. Look up Px  0 on the Cumulative Binomial Table with n  5 and p  .3 ,
and find .16807.
b) At least 3 repaired on the same day: We want Px  3 , actually, the probability of between 3
and 5 successes. But, if we use the .3 column of the table again, we need to know that 3
successes correspond to 2 failures and that 5 successes correspond to 0 failures. We look up
Px  2 on the Cumulative Binomial Table with n  5 and p  .3 , and find .83692.
c) Fewer than 2 repaired on same day: We want Px  2  Px  1 , actually between 0 and one
successes. But, if we use the .3 column of the table again, we need to know that 1 success
corresponds to 4 failures and that 0 successes corresponds to 5 failures. We look up
Px  4  1  Px  3  1  .96922  .03078 .
Instead of answering the above, the Instructor’s Solutions Manual provides the following Excel
table.
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d) According to the Instructor’s Solutions Manual there are two assumptions:
(1) Independence of the repairs, (2) Only two outcomes - repair accomplished same day or repair
not accomplished same day.
e)   np  5.7  3.50,  2  npq  3.50.30  1.05, .   1.05  1.0247.
f) Now p  .8 and n  5. Again, we must think in terms of failures instead of successes. The
probability of failure is now .2.
(a) 5 successes corresponds to 0 failures and, on the Cumulative Binomial Table with
n  5 and p  .2 , we find Px  0  .32768 .
(b) We want Px  3 , actually, the probability of between 3 and 5 successes. But, if we
use the .2 column of the table again, we need to know that 3 successes correspond to 2
failures and that 5 successes correspond to 0 failures. We look up Px  2 on the
Cumulative Binomial Table with n  5 and p  .2 , and find .94208.
(c) We want Px  2  Px  1 , actually between 0 and one successes. But, if we use
the .2 column of the table again, we need to know that 1 success corresponds to 4 failures
and that 0 successes corresponds to 5 failures. We look up Px  4  1  Px  3
 1  .99328  .00672 .
(e)   np  5.8  4.00,  2  npq  4.00.20  0.80, .   0.80  0.8944
g) According to the Instructor’s Solutions Manual, “The larger p is, the more likely it is that
troubles reported on a given day will be repaired on the same day, and the less likely it is that
troubles reported on a given day will not be repaired on the same day.”
Exercise 5.23 [Exercise 5.20 in 8th and 9th]: A student takes a multiple choice exam with 5 questions,
each with four answers a, b, c and d. For each question she picks an answer randomly so that each answer
has an equal chance .25  of being picked. Find the following probabilities: a) five questions correct; b) at
least four questions correct; c) no questions correct and d) no more than two questions correct.
Solution: This is a binomial problem with a sample size of 5 and a probability of success of .25. All
answers are taken from the binomial table.
(a)
P(X = 5) = Px  5  Px  4  1 .99902 = 0.0010
(b)
(c)
(d)
P(X
 4) = 1  Px  3  1  .98437  .0156
P(X = 0) = Px  0 = 0.2373
P(X  2) = 0.8965
Exercise 5.27 [Exercise 5.21 in 8th and 9th]: On one day 20 customers place orders. Each has a 5%
probability of exceeding their credit limit. Find a) the mean and standard deviation of the number of
customers exceeding the limit, b) the probability that no customers will exceed the credit limit c) the
probability that one customer exceeds the credit limit and d) the probability that at least two will exceed the
limit.
Solution: This is a binomial problem with a sample size of 5 and a probability of success of .05. All
answers are copied from the Instructor’s Solutions Manual and are unchecked. If you cannot get them,
please tell me.
(a)
mean = 1
standard deviation = 0.9747
(b)
P(X = 0) = 0.3585
(c)
P(X = 1) = 0.3774
(d)
P(X  2) = 0.2642
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Geometric Distribution Problems
Problem L2:.a. If p  .8 , what is the chance of the first success on the 9th try? The 5th? The first?
b. If p  .8 , what is the mean number of tries to a success? The variance?
c. What is the probability that the first success occurs between the 3 rd and the 7th try?
Solution: The outline gives the formula: P( x)  q x1 p , which tells us the probability that the first success
occurs on try x . The cumulative distribution, which is used more frequently is F x  1  q x . The mean is :

q
1
. and the variance is  2  2 .
p
p
a) If p  .8 P(9)  q 8 p  .28 .8  .000002 , P(5)  q 4 p  .24 .8  .00128 and
P(1)  q 0 p  .20 .8  .8 (Surprise?)
b) The mean is :  
q
1 1
.2
 0.3125 .
  1.25 . and the variance is  2  2 
p .8
p
.82
  0.55902 
c) P3  x  7  Px  7  Px  2  F 7  F 2 The cumulative distribution formula is F x  1  q x , so

 

Px  7  Px  2  1  q 7  1  q 2  q 2  q 7  .22  .27  .0400000  .0000128  .03999 .
Problem L3: a. If a bunch of fair coins are spilled and I pick them up randomly, what is the chance of the
first head on try 1? Try 2? Try 3? If x is the first success, what is the probability that x is at least 3?
b. On the average, how many coins will I have to pick up to get a head?
Solution: a) P( x)  q x1 p and p  .5 P(1)  q 0 p  .50 .5  .5 , P(2)  q 1 p  .51 .5  .25 ,
P(3)  q 2 p  .52 .5  .125 . In order to have our first success on the 3 rd or later try, we must have failures
on the 1st and 2nd try. P( x  3)  q 2  .52  .25 , but remember that F x  1  q x so


P( x  3)  1  P( x  2)  1  F 2  1  1  q 2  q 2  .52  .25 .
b)  
1 1
 2
p .5
Problem L4: a. If p = .01, Use the Chebyshef inequality to give a range in which three quarters of first
successes must lie.
b. Try confirming this using the Cumulative probability for this distribution.
Solution: According to the inequality, if we define a 'tail' of the distribution as those points more than
k  2 standard deviations from the mean, then the proportion of points in the tails will be at most 1 k 2  1 4
and the proportion within 2 standard deviations of the mean will be at least
3
4
. We already know that
q
1 1
.99
 9900 . So   9900  99.4987 . The lower tail

 100 and the variance is  2  2 
p .01
p
.012
ought to begin at   2  100  299.4987   99 , but this is impossible since x can't be below 1. The

upper tail ought to begin at   2  100  299.4987   299 So at least three-quarters of first successes will
occur between the first and the 299th try.
b) F x  1  q x , so Px  299   F 299   1  .01299  1. Actually if we set F x   1  .99  x  .75 ,
then .99  x  .25 . A little fooling around with powers of .99 indicates that x is about 137.
251solnL1 3/29/06 (Open this document in 'Page Layout' view!)
Problem L5: I am a real estate agent who gets an average of 1 customer every three days.
a. If I go away for three days, what is the chance that I miss a customer?
b. If I sit and wait for two days, what is the chance that the first customer doesn't
come in until the third day?
a) p 
1
, so q  2 3 and F x  1  q x . So Px  3  F 3  1  2 3   1  8 27  .70370 .
3
3
b) P( x)  q x1 p , P(3)  q 2 p  2 3 
 3   4 27  .14815 .
2 1