Ela Jackiewicz
Mat 119 Exam #4 Key
In problem #1 just set-up.
1.) The 5-card hand is drawn from an ordinary deck of 52 cards.
a.) (7 points) Compute the probability that exactly two cards are hearts.
C(13.2)*C(39,3)
------= .274
C(52,5)
b.) (7 points) Compute the probability that no more than 4 cards are hearts.
(hint: use opposite event)
1 - P(all 5 hearts) = 1-C(13,5)/C(52,5)
In problem #2 just set-up.
2.) At the New Year's party you find yourself in a company of 5 strangers (so, there
are 6 of you)
a. (7 points) What is the probability that all of you have different birthdays?
P(365,6)/ 3656
b. (7 points) What is the probability that at least 2 of you were born in the same
month? (hint: use opposite event)
1 - P(all born in different months) =1 - P(12,6)/126
3. Assume the probability of having a boy is 51% and the probability of having a
girl is 49%. What is the probability that in a family of five children:
a). (7 points) First 3 children are boys, the rest are girls ?
P(BBBGG)= (.51)3*(.49)2 = .032
b). (7 points) At least one child is a boy?
1 - P(all 5 girls) = 1-(.49)5 = .972
4. U.S. Resident Population, by Race and Age, as of July 1, 1998. Adapted from
[Data is in millions of people, rounded to
the nearest tenth of a million people. "White" includes Hispanic. "Other"
includes American Indian, Eskimo, Aleut, Asian, and Pacific Islander.]
AGE / White / Black / Other / totalUnder age 20 / 61.4 / 12.0 / 4.2 / 77.6
Age 20 - 39 / 63.9 / 10.9 / 4.2 / 79
Age 40 or over / 97.6 / 11.5 / 4.4 / 113.5
total / 222.9 / 34.4 / 12.8 / 270.1
A July 1, 1998 U.S. resident is selected at random. Find the probability that the resident
a). (5 points) Is Under age 20.
77.6/270.1= .287
b). (5 points) Is age 40 or over and Black.
11.5/270.1 =.043
c). (5 points) Is White, given that the person is Age 20-39.
63.9/79 = .81
In problem #5 just set-up.
5. A manufacturer of transformers knows that about 3% of his transformers are
defective. He ships 70 random transformers to one of his clients.
Use the binomial probability model to answer questions a and b.
a). (7 points) Calculate the probability that the customer has received exactly 7
bad transformers.
C(70,7)(.03)7(.97)63 = .00385
b). (7 points) Calculate the probability that the customer has received at least 2
bad transformers. (hint: use opposite event)
1 - P (1 bad or 0 bad transformers)
= 1 -[ C(70,1)(.03)1(.97)69 + C(70,0)(.03)0(.97)70] = .625
6. Probability that Professor Jones is late for his lecture is 0.20. If he is on time, the
probability that he will give a quiz is 0.6. If he is late, the probability that he will
give a quiz is only 0.35.
a). (8 points) Draw a tree diagram (with all probabilities marked) for this
problem.
b). (7 points) One random day, Professor Jones gave a quiz, what is the
probability that he was late that day?
P(late/Q) = [(.2)(.35)] / [(.2)(.35)+(.8)(.6)] = .127
7. Your company is considering purchasing insurance for a loan it made. If the
borrower pays the loan in full, then the company makes $125,000. If the
borrower defaults on the loan, the company can sell the house and make
$30,000. History shows that the probability that the person will default is 0.3.
a.) (6points) What are your company's expected earnings?
E= ($125,000)*(.7) + ($30,000)*(.3) = $96,5000
b.) (6 points) The insurance company will charge you $10,000 for $70,000 worth of insurance that pays only if the borrower defaults. What are the expected earnings if the company buys the insurance?
If client defaults bank's profit is: $30.000+$70,000-$10.000=$90,000
If he does not default, bank's profit is: $125,000-$10,000=$115,000
E= (.3)*($90,000) + (.7)*($115,000) = $107,500
c.) (2 points) Based on parts a and b, should the company buy the insurance?
Expected profit is larger with insurance, so the bank will not lose on average if it will decide to purchase one.