Spring 2009, Math 196, Test 3 Solutions
Problem 1. In the future value formula, if F V = 5000, n = 15, and i = .001, what is P M T ?
Solution. The future value formula is:
FV = P MT ·
(1 + i)n − 1
i
Plugging stuff in and solving for P M T gives us:
(1 + .001)15 − 1
.001
5000 = P M T · 15.11
331.01 = P M T
5000 = P M T ·
Problem 2. Recently, Guaranty Income Life offered an annuity that pays 6.65 percent compounded monthly. If 500 dollars is deposited into this annuity every month, how much is in the
account after 10 years? How much of this is interest?
Solution. Since monthly payments are being made, we know this is either a future value or a
present value problem. Since we are depositing money now to get more money in the future, it is
a future value problem. So we use the future value formula:
FV = P MT ·
(1 + i)n − 1
i
We have:
• F V is unknown
• P M T = 500 (the monthly payments)
• i = .0665/12 = .0055
• n = 12 · 10 = 120 (12 payments a year for 10 years)
Plugging all this in gives us:
(1 + .0055)120 − 1
.0055
F V = 500 · 169.79
F V = 84895.40
F V = 500 ·
After ten years, there is 84, 895.40 dollars in the account. There have been 120 payments of 500
dollars each, so 120 · 500 = 60, 000 has been paid into the account. Anything over that is interest,
so 84895.40 − 60000 = 24895.40 has been made in interest.
Problem 3. In the present value formula, if P M T = 250, n = 25, and i = .025, what is P V ?
Solution. The present value formula is:
P V = P MT ·
1
1 − (1 + i)−n
i
Plugging everything in gives us:
1 − (1 + .025)−25
.025
P V = 250 · 18.42
P V = 4606.09
P V = 250 ·
Problem 4. If you buy a computer directly from the manufacturer for 2,500 dollars and agree
to repay it in 48 equal installments at 1.25 interest per month (so i = .0125) on the unpaid balance,
how much are your monthly payments? How much total interest will you pay?
Solution. Since monthly payments are being made, we know this is either a future value
problem or a present value problem. Since you borrow the money to have it now, it’s a present
value problem, and we use the present value formula:
P V = P MT ·
1 − (1 + i)−n
i
We have:
• P V = 2500 (the amount you are borrowing)
• P M T is unknown
• i = .0125
• n = 48 (the number of payments)
Plugging all this in gives us:
1 − (1 + .0125)−48
.0125
2500 = P M T · 35.93
69.58 = P M T
2500 = P M T ·
Each payment is 69.58 dollars. You make 48 payments, so you pay a total of 48 · 69.58 = 3339.84
dollars. You borrowed 2500 dollars, so anything you paid back above that amount is interest. You
paid 3339.84 − 2500 = 839.84 dollars in interest.
Problem 5. Solve for x and y, if possible:
−6x + 10y = −30
3x − 5y = 10
Solution. Multiply the bottom equation by 2 to get:
−6x + 10y = −30
6x − 10y = 20
Adding the two equations together gives:
0 = −10
2
which is a false statement, and so there are no solutions to these equations.
Problem 6. Solve for x and y, if possible:
9x − 3y = 24
11x + 2y = 1
Solution. Multiply the top equation by 2 and the bottom equation by 3 to get the coefficients
of Y to line up:
18x − 6y = 48
33x + 6y = 3
Add the equations together to get:
51x = 51
and so x = 1. Plugging this back into the first equation gives:
9 · 1 − 3y = 24
−3y = 15
y = −5
Problem 7. List the elements in the following set:
{x|x is a month starting with M}
Solution. All the months starting with M are:
{M arch, M ay}
Problem 8. List the elements in the following set:
{1, 2} ∩ {2, 3, 4}
Solution. The only element in common to the two sets is two, so
{1, 2} ∩ {2, 3, 4} = {2}
Problem 9. Fill out the Venn diagram below, given that n(A) = 45, n(B) = 55, n(A∪B) = 80,
and n(U ) = 100. (See the Venn diagram for Exercises 21-26 in section 7-3.)
Solution. The addition principle is:
n(A) + n(B) − n(A ∪ B) = n(A ∩ B)
So here we have n(A ∩ B) = 45 + 55 − 80 = 20.
Since A is split into two parts, A ∩ B and A ∩ B ′ , we have:
n(A ∩ B) + n(A ∩ B ′ ) = n(A)
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And so 20 + n(A ∩ B ′ ) = 45, and so n(A ∩ B ′ ) = 25.
Since B is split into two parts, A ∩ B and B ∩ A′ , we have:
n(A ∩ B) + n(B ∩ A′ ) = n(B)
And so 20 + n(B ∩ A′ ) = 55, and so n(B ∩ A′ ) = 35.
Finally, since n(U ) = 100 and n(A ∪ B) = 80, there are 20 things outside of A and B.
Problem 10. A particular new car model is available with 5 choices of color, 3 choices of
transmission, 4 types of interior, and 2 types of engine. How many different variations of this model
car are available?
Solution. This is a sequence of choices, so we multiply the number of choices together:
5 · 3 · 4 · 2 = 120
There are 120 different variations.
Problem 11. Permutation or combination?
(a) A book club meets monthly in the home of one of its twelve members. In December the club
selects a host for each meeting of the next year.
(b) Three cards are drawn from a 52-card deck.
Solutions. In (a), the order in which the host is chosen matters, so its a permutation. In (b),
the order in which the cards are drawn doesn’t matter, so it’s a combination.
Problem 12. A jewelry story chain with 8 stores in Georgia, 12 in Florida, and 10 in Alabama
is planning on closing 10 of these stores. The company decides to close 2 stores in Georgia, 5 in
Florida, and 3 in Alabama. In how many ways can this be done?
Solution. This is a two-step problem.
First, there are C8,2 ways to select two stores out of eight in Georgia to close, there are C12,5
ways to select five stores out of twelve in Florida to close, and there are C10,3 ways to select three
stores out of ten in Alabama to close.
Second, the company has to make a sequence of choices: first, which stores to close in Georgia,
then, which stores to close in Florida, and finally which stores to close in Alabama. So we multiply
the number of choices together:
C8,2 · C12,5 · C10,3
is the number of ways they can close these stores.
Problem 13. One card is drawn from a 52 card deck. What’s the probability of drawing a
diamond or a club?
Solution. There are 26 cards that are either diamonds or clubs so the probability of drawing a
diamond or a club is 26/25 = 1/2.
Problem 14. 5 cards are dealt from a standard 52 card deck. What’s the probability of being
dealt three aces and two kings?
Solution. To get a five card hand with three aces and two kings, you have to first choose 3
of the four aces, and there are C4,3 ways to do that, and then you have to choose 2 of the four
kings, and there are C4,2 ways to do that. So the total number of ways to get a five card hand with
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three aces and two kings is C4,3 · C4,2 . There are C52,5 ways to pick five cards from 52, so that’s the
number of five card hands. The probability of getting three aces and two kings is thus:
C4,3 · C4,2
C52,5
Problem 15. From a survey involving 1000 students at a large university, a market research
company found that 750 students owned stereos, 450 owned cars, and 350 owned both cars and
stereos. If a student at the university is selected at random, what is the (empirical) probability
that:
(a) the student owns either a car or a stereo?
(b) the student owns neither a car nor a stereo?
Solution. This is a Venn diagram problem. Since 750 students owned stereos and 350 owned
both cars and stereos, 400 students owned just a stereo and no car. Likewise, since 450 students
owned cars and 350 owned both cars and stereos, 100 students owned a car but not a stereo. So
we have 350 + 400 + 100 = 850 students either owned both, or owned one but not the other. Since
there were 1000 students in the survey, the probability that student picked at random owned either
a car or a stereo (or both) is 850/100 = .85. Since 850 students owned either a car or a stereo (or
both), that leaves 150 students that owned neither, and so the probability that a student picked at
random owned neither a car nor a stereo is 150/1000 = .15.
Problem 16. If the probability is .51 that a candidate wins the election, what is the probability
that he loses?
Solution. Either he wins, or he loses, and so the probability of losing equals 1 − .51 = .49.
Problem 17. An automobile manufacturer produces 37 percent of its cars at plant A. If 5
percent of the cars manufactured at plant A have defective emission control devices, what is the
probability that one of this manufacturer’s cars was manufactured at plant A and has a defective
emission control device?
Solution. A multiplication principle problem. First you pick plant A with 37% probability,
and then you pick a defective car from plant A with 5% probability. And multiply the probabilities:
.37 · .05 = .0185. If you pick a car at random, there is a 1.85% chance it’s a broken car from plant
A.
Problem 18. 2 balls are drawn in succession out of a box containing 2 red and 5 white balls.
(a) Construct a probability tree for this experiment, assuming that the first ball drawn was not
replaced before the second draw.
(b) Find the probability that the second ball drawn was red.
Solution. For the first draw, the probability of drawing a red ball is 2/7 and the probability of
drawing a white ball is 5/7. For the second draw, things are a little more complicated:
1. If the first draw was red, then the probability of getting a red ball on the second draw is 1/6
and the probability of getting a white ball on the second draw is 5/6.
2. If the first draw was white, then the probability of getting a red ball on the second draw is
2/6 and the probability of getting a white ball on the second draw is 4/6.
There are two ways to get a red ball on the second draw:
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1. Draw red and then red, with probability (2/7) · (1/6) = 2/42
2. Draw white and then red, with probability (5/7) · (2/6) = 10/42
Adding these probabilities together gives you a 12/42 probability of drawing a red ball on the second
draw.
Problem 19. After paying 4 dollars to play, a single fair die is rolled and you are paid back
the number of dollars corresponding to the number of dots facing up. For example, if a 5 turns up,
5 dollars is returned to you for a net gain, or payoff, of 1 dollar; if a 1 turns up, 1 dollar is returned
for a net gain of −3 dollars; and so on. What is the expected value of the game?
Solution. We set up the following table:
Roll Probability Payoff Prob times Payoff
1
1/6
−3
−3/6
2
1/6
−2
−2/6
3
1/6
−1
−1/6
4
1/6
0
0
1/6
1
1/6
5
6
1/6
2
2/6
Adding up the numbers in the last column gives us −3/6 = −.50. This is the expected value of
the game: you lose 50 cents.
Problem 20. A five card hand is dealt from a standard 52 card deck. If the hand contains at
least one king, you win 10 dollars; otherwise, you lose 1 dollar. What is the expected value of the
game?
Solution. First, we need to figure out the probability of getting a five card hand with no kings.
There are 48 cards that are not kings, and so there are C48,5 ways to pick five cards that are not
kings. Since there are C52,5 possible five card hands, the probability of getting a five card hand with
no kings is
C48,5
= .658842
C52,5
There’s about a 66% chance of getting a five card hand with no kings.
Every other hand has to have at least one king in it, so there is a 1 − .658842 = .341158
probability of getting a hand with at least one king.
We set up the following table:
Event
Probability Payoff Prob times Payoff
no king
0.658842
−1
−0.658842
at least one king
.341158
10
3.41158
Adding up the numbers in the last column gives us 2.752738 as our expected value; if you play
this game, on average you would win 2.75 dollars.
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