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# Discrete Distributions : Bernoulli, Binomial, Discrete Uniform, Geometric Negative Binomial or Poisson

My main problem is deciding with discrete distribution to use: BERNOULLI, BINOMIAL, DISCRETE UNIFORM, GEOMETRIC NEGATIVE BINOMIAL, OR POISSON. Every time I choose one, it's the wrong one. Is there some way I can easily find out which one to use. Because what I do now is I choose by trial and error ,which takes me a long time but I get the work done sometimes. When it says: the probability that at most or at least sometimes gets me confused because I can sometimes set it up like
P(y>=2) but I don't know what to do from there, use the compliment
or not. Can you please explain how you came to the decision of choosing the discrete distribution you chose. i would greatly appreciate it.
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3.40
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Let Y denote a random variable that has a geometric distribution, with a probability of success on any trial denoted by p.

a) Find P(Y>=2) if p=0.1

Ans: 1-P(Y=1) = 0.9 ,,,,,,, I understand part a). I got correct ans.

b) Find P(Y>4 | Y>2) for general p. Compare this result with the
unconditional probability P(Y>=2).[This property is referred to
as "lack of memory"]

Ans: Since Y>4 is a subset of Y>2, i thought,
P(Y>2)= 1-P(Y=2) = 1-(1-P)*P but answer is wrong
correct ans= (1-p)^2 , why? maybe I wasn't suppose to use
compliment.

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3.54
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Let Y denote a random variable that has a POISON distribution with mean
&#955; =2. Find the following probabilities.
Table &#955; x equals
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2 0 .135
2 1 .406
2 2 .677
2 3 .857
2 4 .947

a) P(Y=4)
Ans: I understand this one, if i didn't i would be in deeper
trouble :-)

b) P(Y>=4)
Ans: I thought 1-P(y=4)- P(y=3) - P(y=2) - P(y=1) = 0.187 ,,
incorrect ans.
correct ans: 0.143
So, I played around with it a while and I found that it's : 1-
P(Y=3): but why. wouldn't
you have to subtract y=2 and 1 as well?
c) P(Y<4)
why is correct answer, only P(X=3)?

d) P(Y>=4 | Y>=2)
I thought about using conditional probability, would that be
correct to use?
Correct Ans: 0.2407
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3.58
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In the article cited in Exercise 3.57
( the national maximum speed limit NMSL of 55 miles per hour was imposed in the U.S. in early 1974. The benefits of this law have been studied by D.B. Kamerud (1983), who reported that the fatality rate for interstate highways with the NMSL in 1975 was apporximately 16 per 10^9 vehicle miles.),
the projected fatality rate for 1975 if the NMSL had not been in effect was 25 per 10^9 vehicle miles. Assume that these conditions had prevailed.

a) Find the probability that at most 15 fatalities occurred in a given
block of 10^9 vehicle miles.

b) Find the probability that at least 20 fatalities occurred in a given
block of 10^9 vehicle miles.

ANS: not sure on what distrubution to use.
Correct Ans: a) 0.022 b) 0.866

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3.64
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The number of imperfections in the weave of a certain textile has a Poisson distribution with a mean of four per square yard.

a) Find the prob. that a 1-square-yard sample will contain at least one
imperfection.
Ans: I understand
1-P(Y=0) = 0.982

b)Find the probability that a 3-square-yard sample will contain at
least one imperfection.
Ans: not sure what to do.
Correct Ans: 1- e^(-12)

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3.66
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The number of bacteria colonies of a certain type in samples of polluted water has a Poisson distribution with a mean of two per cubic centimeter.

a)If four 1-cubic-centimeter samples of this water are independently
selected, find the probability that at least one sample will contain
one or more bacteria colonies.

b)How many 1-cubic-centimeter samples should be selected to establish a
probability of approximately 0.95 of containing at least one bacteria
colony?

Ans: not sure how to set-up and do problem.
Correct Ans:
a) 0.9997 b)2

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3.69
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The number of cars entering a parking lot is a random variable that has a Poisson distribution, with a mean of four per hour. The lot holds only twelve cars.

a)Find the probability that the lot will fill up in the first hour.
(Assume that all cars stay in the lot longer than one hour)

b)Find the probability that fewer than twelve cars will arrive during an
eight-hour day.

Ans: It's a good thing the problem tells you what
distribution it has but still I have
problems setting up problem. I would think P(Y=1)
but it's not that simple since i got answer wrong.

Correct Ans: a) 0.001 b) 0.000017

#### Solution Preview

3.40
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Let Y denote a random variable that has a geometric distribution, with a probability of success on any trial denoted by p.

a) Find P(Y>=2) if p=0.1

Ans: 1-P(Y=1) = 0.9 ,,,,,,, I understand part a). I got correct ans.

b) Find P(Y>4 | Y>2) for general p. Compare this result with the
unconditional probability P(Y >= 2).[This property is referred to
as "lack of memory"]

Ans: Since Y > 4 is a subset of Y > 2,
P(Y>2) = 1 - P(Y<=1) = 1 - P(Y = 0) - P(Y = 1) = 1 - p - (1 - p)*p = 1 - 2p - p2 = (1 - p)2
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3.54
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Let Y denote a random variable that has a POISON distribution with mean
λ =2. Find the following probabilities.
x P(x) Cumulative
0 0.135335 0.135335
1 0.270671 0.406006
2 0.270671 0.676676
3 0.180447 0.857123
4 0.090224 0.947347

a) P(Y = 4) = 0.090224
b) P(Y >= 4) = 1 - P(Y <= 3) = 1 - P(y = 0) - P(y = 1) - P(y = 2) - P(y = 3) = 1 - 0.135 - 0.271 - 0.271 - 0.181 = 0.143
c) P(Y < 4) = P(0) + P(1) + ...

#### Solution Summary

Discrete distributions and probability are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.

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