Transcript Lecture 22 More exercises with answer

Nested Problem and extra
How to solve a probability problem
 1. Read the problem carefully.
 2. Answer the following questions:
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A. What is the random variable of interest in the problem?
B. How can I find the pmf of the r.v.?
1. If this r.v. follows a specific distribution (binomial, Poisson,
HG), find the parameters.
 2. If can not tell the distribution, create the pmf chart.
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C. Whether the problem asks for probability, expected value or
variance?
 3. Check the cheat sheet for corresponding formula
 4. Always remember some tricks: e.g. use complement
Example I
 A corner of the library was on fire and a librarian was
trying to remove as many books from that place as
possible. There are 120 volumes of tax code and 80
textbooks stored in that. After the mission is done, a
count shows that there are 100 books rescued and 60
of them are textbooks. Do you think the librarian had
some selection bias when he/she was rescuing those
books?
Example I
 Following the steps
 1. What is the random variable of interest in this
problem? Is it discrete or continuous?
X=number of textbooks rescued and it is discrete
 2. What probability distribution does this r.v.
follow? What is/are the parameter(s) for that
distribution?
X~HG(200, 100, 80)
 3. Answer the question.
 P(X=60)=80C60*120C40/200C100=4.47268E-09
 Since the probability is very small, there may be some selection bias
involved.
Example II
Peter, the family guy, is fishing at a pond. Two thirds
of the fish in the pond are white bass and one third
are rainbow trout. He decides to stop when he gets
the first trout or 3 white bass. Assuming that he will
get a fish each time trying regardless of the time.
1. Let X be the number of times Peter has to try
before he stops, find the pmf of X.
x
1
2
3
p(x)
1/3
2/9
4/9
Example II
 Let Y be the number of fish Peter may take home.
Find the pmf of Y.
 Same as X.
 If Peter does that every week, on average, how many
fish can he take home each time?
 E(Y)=1*1/3+2*2/9+3*4/9=19/9
Example II
 On the 10th time Peter went fishing, before he got
any fish, he was caught by a policeman. Peter was
told that it is illegal to fish in the pond and a fine will
be imposed based on the number of fish he has got
from the pond. Peter was scared and confessed
EVERYTHING, but he could not remember how
many fish he has got from the pond. The policeman
therefore decided to square the number of fish Peter
got each time and fine him $60 per fish based on
that. How much do you think Peter will pay?
 The money Peter expect to pay is:
 $60*9*E(Y^2)
 Using the formula E(Y^2)=var(Y)+[E(Y)]^2 and
plug in E(Y)=19/9 and Var(Y)=62/9, we have
E(Y^2)=47/9
 Finally, Peter expect to pay 60*9*47/9=$2820
Example III
 Two players, A and B, are playing a card game. They
start with a deck of 26 cards with all clubs and
diamonds removed. A will deal 13 cards at random to
B and A wins if he gets more Ace than B. If both A
and B get one Ace, the one with spade Ace wins. They
repeat the game 10 times. Find the probability that A
wins 6 times.
Assuming each game is independent, the number of
games A wins follows a binomial distribution with
parameters (10 and p)
Example III
 Step 1, find p.
 A wins when he has two Aces or spade Ace, let’s find
the probability of each event separately.
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A. A has two Aces: let X1 be the number of Ace at A’s hand,
then apparently, X1~HG(26, 13, 2), therefore
P(X1=2)=2C2*24C11/26C13=0.24
B. A has only the spade Ace: we can calculate the probability
directly, which is 24C12/26C13 =0.26(note that, we don’t want
to use 25C12 on the top since in that case, it will also include
the possibilities with 2 Aces)
Finally, P(A wins)=0.24+0.26
Example III
 Step 2, find the probability that A wins 6 times.
 Let Y be the number of times A wins and Y~BIN(10, 0.5)
 P(Y=6)=10C6*0.5^10=0.205
What if we want to calculate the probability that A wins 5 times?
Then P(Y=5)=10C5*0.5^10=0.246
Example III
 If they repeat the game 100 times, how many times
do you expect B to win?
 X=number of times B wins and X~BIN(100,0.5)
 E(X)=50
Example IV
 Someone wants to open a store at downtown
Lafayette. He has decided to have his store open
Monday through Saturday but has not decided the
hours yet. He was torn between opening at 8 or 9. He
is willing to open the store at 8 if there are more than
6 customers visiting between 8 and 9 for at least four
days of a week. A quick research told him that on
average, there are about 8 customers visiting a store
in the neighborhood between 8 and 9. What is the
probability that the storekeeper starts his business at
8?
Example IV
 This is also an example of a nested problem.
 The storekeeper’s decision is based on the number
of days from Mon. to Sat. that there are more than
10 customers visiting his store. This number follows
a binomial distribution.
 For a binomial r.v., we need to figure out the two
parameters, n and p to be able to work on it.
 In this case, n=6, Monday through Saturday.
 p depends on the number of customers visiting, that
is a Poisson r.v. with a mean of 8.
Example IV
 X=number of days with more than 6+ customers visiting
between 8 and 9 and X~BIN(6, p), where p is the
probability that 6+ customers visit between 8 and 9 on one
day.
 Need to find p first:
 Y=number of customers visiting between 8 and 9 on one
day and Y~Poi(8)
 p=P(Y>6)=1-P(Y<=6)=
 1-[P(Y=0)+P(Y=1)+P(Y=2)+P(Y=3)+P(Y=4)+P(Y=5)+p(Y=6)]=0.69
 Therefore X~BIN(6, 0.69)
Example IV
 The storekeeper will start at 8 when X>=4, the
probability is then:
 P(X>=4)=P(X=4)+P(X=5)+P(x=6)=0.73.
 Therefore, there is 73% probability that the store
keeper will open the store at 8.
Example IV
 Assume after a week’s trial business, the shop keeper
finally decided to open the store at 9am instead of 8,
then what is the probability that he had 6+
customers visiting between 8 and 9 on only three
days of the week.
 Since the shopkeeper will open the store at 8 if he
has 6+ customers visiting for at least four days of the
week, he not opening the store at 8 means he either
has 0 or 1 or 2 or 3 days with 6+ customers.
 That probability is 1-0.73=0.27
Example IV
 This is a Bayes theorem problem. We know the
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outcome, the store is not opened at 8, and we want to
figure out one of the probability that one of the
possible reasons that happened, there are only 3 days
that 6+ customers visited.
Using Bayes Theorem: the answer is
P(3 days with 6+ customers|not open at 8)
which is P(3 days with 6+ customers| 0 or 1 or 2 or 3
days with 6+ customers)
=P(Y=3)/0.27=0.2/0.27=0.74