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STAT 200
Guided Exercise 3
1. An individual’s genetic makeup is determined by the genes obtained from each
parent. For every genetic trait, each parent posses a gene pair; and each contributes ½ of
the gene pair, with equal probabilities, to their offspring, forming a new gene pair. The
offspring’s traits (eye color, baldness, etc.) come from this new gene pair, where each gene
in this pair posses some characteristic.
For the gene pair that determines eye color, each gene trait may be one of two types,
dominant brown, (B) or recessive blue (b). A person possessing the gene pair BB or Bb has
brown eyes, whereas the gene pair bb produces blue eyes.
a. Suppose both parents of an individual are brown eyed, each with a gene pair Bb. What
is the probability that a randomly selected child of this couple will have blue eyes?
Assuming each combination is equally likely,
only 1 out of 4 will have the bb combination
for Blue eyes. The probability is 1/4 = .25; P
(Blue) = 1/4 = .25
Father
Mother
B
b
B
BB
Bb
b
bB
bb
b. If one parent has brown eyes, type Bb, and the other has blue eyes, what is the
probability that a randomly selected child of this couple will have blue eyes?
I assumed the Father had blue eyes.
Assuming each combination is equally likely,
2 out of 4 will have the bb combination for
Blue eyes. The probability is 2/4 = .5
P(Blue) = 2/4 = .5
Father
Mother
b
b
B
Bb
Bb
b
bb
bb
1 c. Suppose one parent is brown-eyed, type BB. What is the probability that a child has blue
eyes?
I assumed the Mother had BB. It doesn’t matter Father
what the Father had. Assuming each combination is equally likely, 0 out of 4 will have b
b
the bb combination for Blue eyes. The probability is 0/4 = 0 P(Blue) = 0/4 = 0 Mother
B
Bb
Bb
B
Bb
Bb
2 2.
On a standard SAT test, a typical question has five possible answers; A, B, C, D, and E. Only
one answer is correct. If you guess you have a 1 out of 5 or 20% chance of being correct.
a.
What is the probability of not being correct on a question if you randomly guess?
This is the complement of guessing correctly
P(not correct) = 1-.20 = .80
b.
What is the probability of being correct on three questions if you you randomly guess?
Assume independence between questions
P(3 right) = .2*.2*.2 = .008
c.
What is the probability of getting at least one question right on three questions if you
randomly guess?
At least one right means the P(3 right)+ P(2 right)+P(1 right)
P(3 right) = .2*.2*.2 = .008
P(2 right) = (.2*.2*.8)*3 = .096
P(1 right) = (.2*.8*.8)*3 = .384
So, P(at least 1 right) = .008+.096+.384 = .488
In other words, 1-P(none right)
1 - (.8*.8*.8) = 1-.512 = .488
d.
What is the probability of getting all three questions wrong on three questions if you
randomly guess?
P(all 3 wrong) = .8*.8*.8 = .512
e.
Training on how to do better on the SAT test advise that you should guess if you can
eliminate possible answers. Suppose on a question you can eliminate two possible
answers. What is the probability that you are right if you randomly guess your answer
on the remaining items.
This is a Conditional Probability: the probability of a correct guess given you eliminate two
answers.
Now we have a 1/3 = .333 chance of guessing the answer correctly.
3 3.
The table below gives the results of a study of 452 individuals enrolled in a methadone
treatment program. The number of positive HIV tests and negative HIV tests are provides separately for
people with different durations of intravenous drug use. Row and column totals are also provided.
Intravenous Drug
Use
HIV Test Result
Tested Positive
Tested Negative
Total
Never
23
53
76
Less than 4 years
77
141
218
At least 4 years
78
80
158
Total
178
274
452
a.
What is the probability that a randomly selected subject in the program will test positively for HIV? P(HIV) = 178/452 = .3938 b.
What is the probability that a randomly selected subject in the program will have had at least 4
years of drug use and test positively for HIV?
P(least 4 years and HIV) = 78/452 = .1726
c.
Given that a randomly selected subject has had at least 4 years of drug use, what is the probability
that he/she will have a positive HIV test?
P(HIV|at least 4 years) = 78/158 = .4937
d.
Are the events “test positively for HIV” and “at least 4 years of drug use” mutually exclusive?
Explain.
No, they both can ocurr at the same time.
e. Are the events “test positively for HIV” and “at least 4 years of drug use” independent? Why?
If they are independent, it means that that P(HIV and at least 4 years) = P(HIV) * P(at least 4 years)
P(HIV) = 178/452 = .3938
P(at least 4 years) = 158/452 = .3496
P(HIV and at least 4 years) = 78/452 = .1726
.1726 does not equal P(HIV) * P(at least 4 years)= .3938*.3496 = .1376
So they are dependent
f. Calculate the following odds and odds ratio for testing positive for HIV.
(1.) Odds of testing positive versus negative for persons with at least 4 years of intravenous drug
use.
4 years of intravenous drug use. 78/80 = .975
(2.) Odds of testing positive versus negative for persons with no intravenous drug use.
no intravenous drug use. 23/53 = .434
(3.) Odds ratio of (1) versus (2). Explain this in words.
.975/.434 = 2.247, those with 4 or more years of drug use were 2.2 times more likely to contract HIV
4 4. A fast food restaurant has determined that the chance a customer will order a soft drink is .90.
The chance that a customer will order a hamburger is .6, and the chance for ordering french fries
is .5.
a. If a customer places an order, what is the probability that the order will include a soft drink and
no fries if these two events are independent?
Let S stand for Soft Drink, with P(S) = .9
Let F stand for French Fries, with P(F) = .5 and the compliment of F, Fc = 1-.5 = .5
The probability formulas state that: P(S ∩ Fc) = P(S) * P(Fc|S). But, if the two events are
independent of each other, then the probability of their intersection simplifies to:
P(S ∩ Fc) = P(S) * P(Fc)
To solve this we simply multiply through the probabilities given to us: .9 x .5 = .4
b. The restaurant has also determined that if a customer orders a hamburger the chance the
customer will also order fries is .8. Determine the probability that the order will include a
hamburger and fries.
The problem states that the P(F|H) = .8 We are told what the conditional probability of ordering
French Fries is .8, given the customer orders a hamburger. So, using the formula for the
probability of an intersection, P(H ∩ F) = P(H) * P(F|H)
We can solve this problem as: P(H ∩ F) = .6 x .8 = .48
ANSWER: the probability that the order will include a hamburger and fries is equal to .48. This
is based on a conditional probability - the two events are not independent. 5 5. No diagnostic tests are infallible, so imagine that the probability is 0.95 that a certain test will diagnose a
diabetic correctly as being diabetic, and it is 0.05 that it will diagnose a person who is not diabetic as
being diabetic. It is known that roughly 10% if the population is diabetic.
What is the probability that a person diagnosed as being diabetic actually is diabetic?
Hint: This is a use Bayes’ theorem problem, which we did not cover in the lectures. There is another
way to handle this problem – mack a mock 2 by 2 table of the data based on the information you already
know. Once the table is complete, you can solve for the conditional probability. Since some of the
probabilities are small, I would sugget you make a table that is based on 100,000 people. I have started
the table for you.
Test Results
Diabetes Status
Diabetic
Not Diabetic
Diabetic
9500
500
10,000
Not Diabetic
4500
85500
90,000
14000
86000
100,000
This is a conditional probability.
Given you are diagnosed as being diabetic, what is the probability that you actually are diabetic?
P(D|Test says D) = 9500/14,000 = .6786
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