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STAT 408 Spring 2009
Homework
(due Friday, February 13, by 3:00 p.m.)
Be sure to show all your work; your partial credit might depend on it.
No credit will be given without supporting work.
1. A bank classifies borrowers as "high risk" or "low risk," and 16% of its loans are
made to those in the "high risk" category. Of all the bank's loans, 5% are in default. It is also known that 40% of the loans in default are to high-risk borrowers.
P(High risk) = 0.16, P(Default) = 0.05, P(High risk | Default) = 0.40.
a) What is the probability that a randomly selected loan is in default and issued to a
high-risk borrower?
P(Default ∩ High risk) = P(Default) ⋅⋅⋅⋅ P(High risk | Default) = 0.05 ⋅⋅⋅⋅ 0.40 = 0..
High Risk Low Risk
Default 0.02 0.03 0.
Default ' 0.14^ 0.81^ 0.
b) What is the probability that a loan will default, given that it is issued to a high-risk
borrower?
P(Default | High risk) =
P(Highrisk)
P(Default Highrisk)
c) What is the probability that a randomly selected loan is either in default or issued to
a high-risk borrower, or both?
P(Default ∪ High risk) = P(Default) + P(High risk) – P(Default ∩ High risk)
d) A loan is being issued to a borrower who is not high-risk. What is the probability
that this loan will default?
P(High risk)
P(Default High risk ) P( Default High risk ) '
' | ' = =
e) Are events {a randomly selected loan is in default} and {a randomly selected
loan is issued to a high-risk borrower} independent? Justify your answer.
P(High risk | Default) = 0.40. P(High risk) = 0.16.
Since P(High risk | Default) ≠ P(High risk), {Default} and {High risk} are
NOT independent.
OR
P(Default | High risk) = 0.125. P(Default) = 0.05.
Since P(Default | High risk) ≠ P(Default), {Default} and {High risk} are
NOT independent.
OR
P(Default ∩ High risk) = 0.02.
P(Default) × P(High risk) = 0.05 × 0.16 = 0.008.
Since P(Default ∩ High risk) ≠ P(Default) × P(High risk), {Default} and {High risk}
are NOT independent.
2. At Initech , 50% of all employees surf the Internet during work hours. 20% of the
employees surf the Internet and play Solitaire during work hours. It is also known
that 60% of the employees either surf the Internet or play Solitaire (or both) during
work hours.
P( Internet ) = 0.50, P( Internet ∩ Solitaire ) = 0.20,
P( Internet ∪ Solitaire ) = 0.60.
a) What proportion of the employees play Solitaire during work hours?
P( Internet ∪ Solitaire ) = P( Internet ) + P( Solitaire ) – P( Internet ∩ Solitaire )
0.60 = 0.50 + P( Solitaire ) – 0.
P( Solitaire ) = 0..
3. An automobile insurance company classifies each driver as a good risk, a medium
risk, or a poor risk. Of those currently insured, 30% are good risks, 50% are medium risks, and 20% are poor risks. In any given year, the probability that a driver will have a traffic accident is 0.1 for a good risk, 0.3 for a medium risk, and 0.5 for a poor risk.
P( Good ) = 0.30, P( Accident | Good ) = 0.10.
P( Medium ) = 0.50, P( Accident | Medium ) = 0.30.
P( Poor ) = 0.20, P( Accident | Poor ) = 0.50.
a) What is the probability that a randomly selected driver insured by this company
had a traffic accident during 2008?
Accident No Accident
Good
0.03 0.27^ 0.30^ P( Accident^ |^ Good ) = 0.10.
Medium
0.50 ⋅ 0. 0.15 0.35^ 0.50^ P( Accident^ |^ Medium ) = 0.30.
Poor
0.10 0.20 P( Accident | Poor ) = 0.50.
0.28 0.72^ 1.
b) If a randomly selected driver insured by this company had a traffic accident
during 2008, what is the probability that the driver is actually a poor risk?
P( Poor | Accident ) =
c) If a randomly selected driver insured by this company did not have a traffic
accident during 2008, what is the probability that the driver is actually a good risk?
P( Good | No Accident ) =
d) Suppose a driver insured by this company is not a poor risk. What is the
probability that the driver had a traffic accident during 2008?
P( Accident | Poor ' ) =
e) The company announced that it will raise the insurance premiums for the drivers
who either are poor risks or had a traffic accident during 2008, or both. What proportion of customers would have their premiums raised?
P( Poor ∪ Accident ) = P( Poor ) + P( Accident ) – P( Poor ∩ Accident )
f) Are events {a randomly selected driver is a medium risk} and {a randomly
selected driver had a traffic accident during 2008} independent? Justify your answer.
P( Accident | Medium ) = 0.30. P( Accident ) = 0.28.
Since P( Accident | Medium ) ≠ P( Accident ), {Medium} and {Accident} are
NOT independent.
OR
P( Medium | Accident ) = 0.15 / 0.28 ≈ 0.5357. P( Medium ) = 0.50.
Since P( Medium | Accident ) ≠ P( Medium ), {Medium} and {Accident} are
NOT independent.
OR
P( Medium ∩ Accident ) = 0.15.
P( Medium ) × P( Accident ) = 0.50 × 0.28 = 0.14.
Since P( Medium ∩ Accident ) ≠ P( Medium ) × P( Accident ), {Medium} and
{Accident} are NOT independent.
g) Are events {a randomly selected driver is a medium risk} and {a randomly
selected driver had a traffic accident during 2008} mutually exclusive? Justify your answer.
Since P( Medium ∩ Accident ) ≠ 0, {Medium} and {Accident} are
NOT mutually exclusive.
P( M M F ) + P( M F M ) + P( M F F ) + P( F M M ) + P( F M F ) + P( F F M )
OR
{ 2 F and 1 M } = { M F F , F M F , F F M }
{ 1 F and 2 M } = { M M F , M F M , F M M }
P( 2 F and 1 M ) + P( 1 F and 2 M ) =
25 3
C
C C
C
C ⋅ C ⋅
5. A family that owns two automobiles is selected at random. Suppose that the
probability that the older car is American is 0.70, the probability that the newer car is American is 0.50, and the probability that both the older and the newer cars are American is 0.40.
New American New Foreign
Old American (^) 0.40 0.30 (^) 0.
Old Foreign 0.10 0.20 0.
a) Find the probability that at least one car is American (i.e. that either the older car
or the newer car, or both cars are American).
P( OA ∪ NA ) = P( OA ) + P( NA ) – P( OA ∩ NA ) = 0.70 + 0.50 – 0.40 = 0..
b) Find the probability that neither car is American.
P( OF ∩ NF ) = 1 – P( OA ∪ NA ) = 1 – 0.80 = 0..
c) Suppose that the older car is American. What is the probability that the newer car
is also American?
P( NA | OA ) =
P(OA)
P( OA NA)
=^4 / 7.
d) What is the probability that the older car is American, given that the newer car is
American?
P( OA | NA ) =
P(NA)
P( OA NA)
e) Are events { the older car is American } and { the newer car is American }
independent? Justify your answer.
P( OA ) ⋅ P( NA ) = 0.70 ⋅ 0.50 ≠ 0.40 = P( OA ∩ NA ). NOT independent.
7. Jack, Mike and Tom are roommates, and every Sunday night they split a large
pizza for dinner. When there is only one slice left, the probability that Jack wants it is 0.40, the probability that Mike wants it is 0.35, and the probability that Tom wants it is 0.25. Suppose that whether or not each one of them will want the last slice is independent of the other two.
( Jack ) = 0.40, P( Jack ' ) = 0.60,
( Mike ) = 0.35, P( Mike ' ) = 0.65,
( Tom ) = 0.25, P( Tom ' ) = 0.75.
a) What is the probability that only one of the roommates will want the last slice?
only Jack Jack Mike ' Tom ' 0.40 × 0.65 × 0.75 = 0.1950,
only Mike Jack ' Mike Tom ' 0.60 × 0.35 × 0.75 = 0.1575,
only Tom Jack ' Mike ' Tom 0.60 × 0.65 × 0.25 = 0.0975.
P( only one wants the last slice ) = P( only Jack or only Mike or only Tom )
= P( only Jack ) + P( only Mike ) + P( only Tom )
b) What is the probability that at least one of the roommates will want the last slice?
P( at least one wants the last slice ) = 1 – P( no one wants the last slice )
= 1 – P( Jack ' ∩ Mike ' ∩ Tom ' )
= 1 – 0.60 × 0.65 × 0.
c) What is the probability that at most one of the roommates will want the last slice?
P( at most one wants the last slice )
= P( only one wants the last slice ) + P( no one wants the last slice )
From the textbook:
1.4-
Let A = { 3 or 4 kings }, B = { 2, 3, or 4 kings }.
P( A (^) | B ) = N(B)
N(A)
P(B)
P(A)
P(B)
P( A B)
^ +
^ +
^ +
1.4-
P ( 6th S ) =
^ ×
^ ×
1 S, 4 F
first 5 : P 6thS 1 S, 4 F
first 5 : P 0 S, 5 F
first 5 : P 6thS 0 S, 5 F
first 5 : P
1.3-
(a) 351 , 325
(b) 1 , 236 , 664
