Suggested Problems - Chapter 3 - Discrete Random
Variables
5.
Problem 3.16 in Barnes
7.
Problem 3.21 in Barnes
11.
Problem 3.42 in Barnes
12.
Problem C3.4 in Barnes
13.
Problem C3.5 in Barnes
14.
Problem C3.6 in Barnes
3.7 The rhino
charging at you will require 3 hits from your .22 caliber pistol before being
sufficiently discouraged to go elsewhere. Under the circumstances, your
shooting ability will yield a .42 probability of a hit. Assuming you will stop shooting after you
have achieved 3 hits, what is the probability that 7 or fewer shots will be
required?
What distribution is this problem describing?
Negative binomial - number of Bernoulli trials to get r
successes
P(x) = r+x-1 C x pr qx
r = 3
p = .42
q = .58
P(success) = sum from x=0 to 4 = 3+x-1 C 3
.423 .58x
P(success) = 3+0-1 C 0 .423
.580 + 3+1-1 C 1 .423 .581 + 3+2-1 C 2 .423
.582 + 3+3-1 C 3 .423 .583
+ 3+4-1 C 4 .423 .584 = .6229
Using the Stat software from Barnes:
F(x < 4) = .6229
3.8 A study of
Prussian horse-cavalry soldiers noted that the probability of such a soldier
being killed by the kick of a horse was .5% per year. In a group of 5000 such soldiers, how likely are more than 30 to
be killed in one year?
What distribution is this problem describing?
Poisson distribution - The number of event occurrences
during a specified period of time
P(x) = (e-m * m x) / x!
m = np = 5000(.005) = 25
p(x> 30) = 1 - p(x < 30)
Using Table A.2, pp.362-363 from Barnes:
3.14 A shelf
reports 20 ceramic resistors. Three of
the resistors are faulty. If 6
resistors are sampled from the shelf with replacement, what is the probability
that 4 or more of the sample will be faulty? Answer the same question if the
resistors are sampled without replacement.
What distribution is this problem describing?
Binomial distribution - the number of successes in n
Bernoulli trials
P(X = x)
= p(x) = nCx * px * q n-x
= [(n! / x!(n - x)!)] * px * q n-x
N = 20 resistors
a = 3 faulty resistors
n = 6 sample size
p (failure) = 3/20 = .15
Sampling with replacement:
Using Table A.1, pp.352 from Barnes:
p(x > 4) = 1 - F(3) = 1 - .994 = .0059
Sampling Without replacement:
p(x > 4) = 0 since only 3 resistors are faulty
3.15 Tru-Round
Piston Rings, Inc. has an average of .5% defective parts from its GMC
line. As a part of their quality
control program, they sample 40 units from each shipping box of 500 units. What is the probability that a sample will
contain 2, 3, or 4 defective units?
What distribution is this problem describing?
Binomial distribution - the number of successes in n
Bernoulli trials
P(X = x)
= p(x) = nCx * px
* q n-x = [(n! / x!(n - x)!)] * px * q n-x
N = 500 resistors
a = 2, 3 or 4 faulty resistors
n = 40 sample size
p (failure) = .005
Using Table A.1, pp.352 from Barnes:
p(x = 2, 3, or 4) = F(4) - F(1) = 1 - .98281 = .01719
Using the Poisson approximation to the binomial since n >
20 and p < .05:
m = np = 40(.005) = .2
Using the Stat software from Barnes:
p(x = 2, 3, or 4) = p(x=2) + p(x=3) + p(x=4) =
p(x = 2, 3, or 4) = .01637 + .00109 + .00005 = .01751
Solving by hand:
p(x=2) = nCx
* px * q n-x = 40C2
* p2 * q 40-2 = [(40! /
2!(40 - 2)!)] * (.005)2 * (.995) 38 = .016
p(x=3) = nCx
* px * q n-x = 40C3
* p3 * q 40-3 = [(40! /
3!(40 - 3)!)] * (.005)3 * (.995) 37
= .001
p(x=4) = nCx
* px * q n-x = 40C4
* p4 * q 40-4 = [(40! /
4!(40 - 4)!)] * (.005)4 * (.995) 36 = .000048
p(x = 2, 3, or 4) = .016 + .001 + .000048 = .017
3.17 The
probability that a B-29 can successfully avoid the German antiaircraft fore and
than make a successful bombing run on the Rhine Bridge is .578. If 3 successful bombing runs are required to
knock out the bridge, how many B-29's should be readied for tonight's raid?
What distribution is this problem describing?
Negative binomial - number of Bernoulli trials to get r
successes
P(x) = r+x-1 C x pr qx
r = 3
p = .578
q = .422
What is the acceptable probability? Let's use 90%
Using the Stat software from Barnes:
F(x < 4) = .8809
F(x < 5) = .9351
Thus we need 5 planes to get 3 hits with 90% confidence.
3.28 In a group
of 32 seals, 12 have blue eyes. How
likely is it that, in a random sample of 7 of the 32 seals, fewer than 4 will
have blue eyes?
What distribution is this problem describing?
Hypergeometric distribution = The number of successes in a
sample of size of n
P(x) = [(aCx)(N-a C n-x)]
NCn = 0,1,2….
Where p = a/N = initial proportion of successes
N = 32 seals
a = 12 seals with blue eyes
n = 7 sample size
Using the Stat software from Barnes:
p(x < 4 blue eyes) = P(x < 3) = .78191
or
p(x < 4 blue eyes) = p(x = 0) + p(x = 1) + p(x = 2) + p(x
= 3) = .78191
3.37 A salesman
makes a sale to a customer with probability of .3. If he must sell 3 items, what is the probability that doing so
will take fewer than 6 customers? Exactly 3 customers?
What distribution is this problem describing?
Negative binomial - number of Bernoulli trials to get r
successes
P(x) = r+x-1 C x pr qx
r = 3
p = .3
q = .7
x = 6
Using the Stat software from Barnes:
P(x < 2) = .163
Exactly 3 customers?
r = 3
p = .3
q = .7
x = 3
Using the Stat software from Barnes:
P(x = 3) = .027
3.38 Thirty percent
of the applicants for a job have advanced computer training. If applicants are randomly selected for
interviewing, what is the probability that the first applicant with such
training is the fifth person interviewed?
What distribution is this problem describing?
Geometric - The number of failures prior to the first
success in a sequence of Bernoulli trials.
P(x)=p * (1-p)x x
= 0,1,2….
p = .3
p(x = 4) = (.30) (1 - .30)4 = .072
What is the probability that the third applicant is found on
the tenth interview?
What distribution is this problem describing?
Negative binomial - number of Bernoulli trials to get r
successes
P(x) = r+x-1 C x pr qx
r = 3
p = .3
q = .7
x = 7
P(x) = 3+7-1 C 7 (.3)3 (.7)7
= 36 (.027)(.08235) = .08