Download Solutions to Practice Exam 2

Solutions to Practice Exam 2
Math 203
February 17, 2012
1. You get bored during a long shift of student work at the entrance to Anderson Commons, and you start to wonder about the names of Concordia students. So you start
keeping track. After 2000 students pass through your line, so far you have counted 414
names ending in “-son,” along with 53 Sarahs and Saras, of whom only 8 had a last
name ending in “-son.”
(a) What is the empirical probability that a Concordia student’s last name ends in
414
= 0.207.
“-son”? Solution. The empirical probability is
2000
(b) If a Concordia student’s last name ends in “-son,” what is the empirical probability
that the student’s first name is Sarah or Sara?
8
Solution. The answer is
≈ 0.0193.
414
2. Suppose you draw two cards at random from a standard deck.
(a) What is the probability that the cards are a pair?
Solution. Every drawing of two cards is equally likely, so we can divide the
number of “good” draws by the total number of draws, which is C(52, 2). If we
want to get a pair, we first choose which rank (aces, sevens, etc.) we want in our
pair (13 choices) and then choose which 2 of the 4 cards in that rank to get. So
the number of pairs in the deck is 13 · C(4, 2). Hence the probability of drawing
78
13 · C(4, 2)
=
≈ 0.0588.
a pair is
C(52, 2)
1326
(b) What is the probability that at least one of the cards is an eight?
Solution. Our denominator is the same as above, but for the numerator there
are two possibilities. Either both cards are eights—C(4, 2) possibilities—or else
just one card is an eight, in which case we can choose that eight in 4 ways and
198
C(4, 2) + 4 · 48
=
≈ 0.1493.
the non-eight in 48 ways. So our answer is
C(52, 2)
1326
(c) What is the probability that the cards add up to 15? Use cribbage rules, where
10, Jack, Queen, and King are all worth 10, but an ace is worth only 1.
Solution. There are three possibilities: either we have a 7 and an 8, or a 6
and a 9, or a 5 and a card worth 10 points. For a 7-8 combination, there are 4
ways to choose the 7 and 4 ways to choose the 8, for 42 ways in all. For a 6-9
combination, there are also 42 ways. But for a combination involving a five, the
other card can be a 10, J, Q, or K; so there are 4 ways to choose the five but 16
ways to choose the other card, for a total of 4 · 16 ways. Hence the probability is
42 + 42 + 4 · 16
96
=
≈ 0.0724.
1326
1326
Practice Exam 2 Solutions
Page 2 of 3
Math 203, Spring 2012
3. Consider a sample space with events E and F , in which p(E) = 0.7 and p(F ) = 0.5.
(a) Can E and F be mutually exclusive events? Explain your answer.
Solution. If E and F were mutually exclusive (a.k.a. disjoint) events, then we
would have p(E ∪ F ) = p(E) + p(F ) = 0.7 + 0.5 = 1.2. But no probability can
ever be greater than one! Therefore E and F cannot be mutually exclusive.
(b) If E and F are independent events, compute p(E ∩ F ) and p(E ∪ F ).
Solution. We recall that if E and F are independent, then p(E ∩F ) = p(E)p(F ).
Thus p(E ∩ F ) = p(E)p(F ) = 0.7 · 0.5 = 0.35. Once we’ve found that, we can
also compute that p(E ∪ F ) = p(E) + p(F ) − p(E ∩ F ) = 0.7 + 0.5 − 0.35 = 0.85.
(c) If p(E ∪ F ) = .9, what is p(E|F )?
Solution. Since 0.9 = p(E ∪ F ) = p(E) + p(F ) − p(E ∩ F ) = 0.7 + 0.5 − p(E ∩ F ),
we see that p(E ∩ F ) must be 0.3. Then applying our formula,
p(E|F ) =
0.3
3
p(E ∩ F )
=
= = 0.6.
p(F )
0.5
5
4. To play the card game called euchre, you only use the 9, 10, J, Q, K, and A of each of
the four suits.
(a) You draw one card at random from a euchre deck. What is the probability of
getting either a black card or1 an ace?
Solution. There are 6·4 = 24 cards in the deck, of which 12 are black, 4 are aces,
and 2 are black aces. Let B denote black and A denote ace; then the probability
of getting a black card or an ace is
Pr(B ∪ A) = Pr(B) + Pr(A) − Pr(B ∩ A) =
4
2
14
7
12
+
−
=
=
≈ 0.5833
24 24 24
24
12
(b) You draw two cards at random from a euchre deck. What is the probability that
the two cards are of the same suit?
Solution. There are C(24, 2) = 276 ways to draw two cards from the deck total.
If we want two cards to be of the same suit, we can choose the suit and then
choose the two cards, so there are C(4, 1)C(6, 2) = 60 ways to do that. Thus the
probability is
60
5
=
≈ 0.2174.
276
23
(c) Your brother shuffles the deck thoroughly and then deals you the top card. Before
he did so, you happened to notice that the bottom card is the jack of diamonds.
What is the probability that he deals you a jack?
Solution. There are 23 cards not on the bottom of the deck, and all of them are
3
equally likely to be on the top. Three of them are jacks, so the answer is
.
23
1
or both, of course!
Page 3 of 3
Practice Exam 2 Solutions
Math 203, Spring 2012
5. A machine arranges a white ball (W), a black ball (B), and a red ball (R) in a row in
a random order.
(a) What is the sample space S for this experiment? List all its elements.
Solution. S = {W BR, W RB, BW R, BRW, RW B, RBW }.
(b) List the elements of the event E = “the white ball is in the middle.”
Solution. E = {RW B, BW R}.
(c) List the elements of the event F = “the red ball is next to the black ball.”
Solution. F = {W BR, W RB, BRW, RBW }
(d) Describe the event E ∪ F in terms of the sample space.
Solution. E ∪ F = {W BR, W RB, BW R, BRW, RW B, RBW } = S.
6. You are with a group of 10 friends, of whom 4 are business majors, 3 are history
majors, and 3 are math majors. Suppose that in general, 60% of history majors like
the movie A Beautiful Mind, 80% of math majors like it, and 50% of business majors
like that movie. If someone in your group of friends likes A Beautiful Mind, what is
the probability that he is a math major?
Solution.
B
L
0.20
0.
4
5
0.
0.3
0.5
H
0.6
D
0.20
L
0.18
Key:
B business major
H history major
M math major
L likes A Beautiful Mind
D doesn’t like A Beautiful Mind
4
0.
Hence
p(M ∩ L)
p(L)
0.24
=
0.20 + 0.18 + 0.24
24
=
62
12
=
≈ 0.3871.
31
3
0.
p(M |L) =
M
0.8
0.12
L
0.24
D
0.06
2
0.
D