Probability
Math 421
Test 1
October 7, 2013
Dr. Pendergrass
On my honor, I have neither given nor received any aid on this work, nor am I aware of any
breach of the Honor Code that I shall not immediately report.
Pledged:
Print Name:
Do these problems in your homework notebook, according to the same guidelines used for homework. You
may use your class notes, and the lecture slides and notes posted on our class web-page. No other resources
are allowed. This test is due at the beginning of class on Monday, October 14, 2013.
1. Events A and B in a probability space have the following probabilities:
P (A) = 0.60,
P (B) = 0.55,
P (A ∩ B c ) = 0.30
(a) Find P (A ∪ B).
(b) Find P (B|A).
(c) Find P (Ac ∪ B c ).
(d) Find P (Ac ∩ B c ).
(e) Are A and B independent?
Solution:
(a) P (A ∪ B) = 0.85
(b) P (B|A) = 0.50
(c) P (Ac ∪ B c ) = 0.70
(d) P (Ac ∩ B c ) = 0.15
(e) No.
2. An all-seeing, all-knowing genie informs you that the Washington Nationals baseball team will go 33-48 at
home next season (i.e. 33 wins, 48 losses for games played at their home stadium). If you go to 19 home
games next year chosen at random, what is the probability that the Nationals lose each one?
Solution:
33 48
0
19 =
81
19
2568691
≈ 7.6217 × 10−6
337022009415
3. A seven-card hand is dealt from a well-shuffled deck of 52 ordinary playing cards. Find the probabilities
of the following events:
(a) P (four of a kind, and three of another kind).
(b) P (three of a kind, and two pairs).
(c) P (at least one ace).
(d) P (straight). (Note: a straight is 7 cards in sequence. An ace can be either high or low in a straight.)
Solution:
13 4
(a)
1
12
1
4
4
52
7
3
=
3
≈ 4.6642 × 10−6
643195
Page 1 of 5
Question 3 continues on next page. . .
Continuing question 3...
(b)
3
42
12
2
52
7
4
13
1
48
7
52
7
(c) 1 −
Test 1
=
2
=
Dr. Pendergrass
594
≈ 9.2351 × 10−4
643195
3478
≈ 0.4496
7735
7−4
8
·
4
8192
18
≈ 9.7972 × 10−4 , or if you exclude the straight flushes,
=
≈
52
8361535
18377
7
7
9.7949 × 10−4
8 · 47
(d) 52 =
4. In medical testing, a false positive occurs when the test indicates that the disease is present, when in fact
no disease is present. A false negative occurs when the test indicates that no disease is present, when in fact
disease is present. Let D be the event that disease is present, + be the event that the test is positive, and −
be the event that the test is negative. Suppose that a certain test has false positive rate P (+|Dc ) = 0.001,
and false negative rate P (−|D) = 0.0005. Suppose also that the prevalence of the disease is relatively rare:
P (D) = 10−6 .
(a) If the test comes back positive, what is the probability that the patient has the disease?
(b) If the test comes back negative, what is the probability that the patient does not have the disease?
Solution:
Note that P (+|D) = 1 − P (−|D) = 0.9995, and P (−|Dc ) = 1 − P (+|Dc ) = 0.999.
(a) By Bayes Theorem
P (D) P (+|D)
P (D) P (+|D) + P (Dc ) P (+|Dc )
10−6 · 0.9995
= −6
10 · 0.9995 + (1 − 10−6 ) · 0.001
≈ 9.985 × 10−4
P (D|+) =
(b) Again by Bayes Theorem
P (Dc ) P (−|Dc )
P (D) P (−|D) + P (Dc ) P (−|Dc )
(1 − 10−6 ) · 0.999
= −6
10 · 0.0005 + (1 − 10−6 ) · 0.999
≈ 1 − 5.005 × 10−10
P (Dc |−) =
5. A bag of candy contains 6 Reese’s Cups, and 9 Snickers bars. Roll a fair six-sided die, and if the outcome
is k, draw k pieces of candy out of the bag.
(a) Find the probability that you get at least one Reese’s Cup.
(b) Let Y be the number of Reese’s Cups you get. Find the probability mass function for Y .
Solution:
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Question 5 continues on next page. . .
Continuing question 5...
Test 1
Dr. Pendergrass
(a) Let X be the result of the die toss, and condition on the value of X:
P (Y = 0) =
6
X
P (X = j) P (Y = 0|X = j)
j=1
6
1X
=
6
6
0
9
j=1
=
j
15
j
2133
10010
Hence, P (Y > 0) = 1 − 2133/10010 ≈ 0.7869.
(b) For k ∈ {1, 2, 3, 4, 5, 6} we have
P (Y = k) =
6
X
P (X = j) P (Y = k|X = j)
j=k
6
1X
=
6
6
k
j=k
9
j−k
15
j
The mass function for Y is tabulated below:
k
0
1
2
3
4
5
6
P (Y = k)
2133
10010
416
1155
817
3003
376
3003
82
3003
32
15015
1
30030
approx.
0.2131
0.3602
0.2721
0.1252
0.0273
0.0021
3.33 × 10−5
6. Urn I contains a one-dollar bill, Urn II contains two one-dollar bills and a ten-dollar bill, and Urn III
contains three one-dollar bills, two ten-dollar bills, and a hundred-dollar bill. But the urns aren’t labeled,
so you don’t know which is which. Choose an urn at random, and then pick a bill at random from that
urn. Let X be the value of the bill you get.
(a) Find the probability mass function for X.
(b) Suppose that X = 1. What is the probability that the bill came from Urn I?
Solution:
(a) Let W be the number of the selected urn, and condition on W :
P (X = x) =
3
X
P (W = w) P (X = x|W = w)
w=1
3
X
1
P (X = x|W = w)
3
w=1

1
2
3

if x = 1
3 1 + 3 + 6
1
1
2
= 3 0+ 3 + 6
if x = 10

1
1
if x = 100
3 0+0+ 6

13

if x = 1
 18
4
= 18
if x = 10

1
if x = 100
18
=
Page 3 of 5
Question 6 continues on next page. . .
Continuing question 6...
Test 1
Dr. Pendergrass
(b)
P (W = 1, X = 1)
P (X = 1)
P (W = 1) P (X = 1|W = 1)
=
P (X = 1)
1
·1
= 313
P (W = 1|X = 1) =
18
6
=
13
7. Compute the probability that a 5-card poker hand is void in at least one suit.
Solution:
Consider the complementary event that a 5-card hand has at least one card from each suit. Then it
must have exactly two cards from one suit, and exactly one card from each of the remaining suits. The
probability of this is
4 13 13 3
2197
1
2
1 =
52
8330
5
So the probability of being void in at least on suit is
4
1
1−
13 13 3
2
1
52
5
=
6133
≈ 0.7363
8330
Alternate Solution: Number the suites, for example as follows: 1 ↔ Hearts, 2 ↔ Clubs, 3 ↔ Spades,
4 ↔ Diamonds. Let Ei be the event that the hand is void in suite i, for i = 1, 2, 3, and 4. We need to
find P ∪4i=1 Ei . We will use inclusion/exclusion. Note that
P (Ei ) =
39
5
52 ,
5
1≤i≤4
P (Ei Ej ) =
26
5
52 ,
5
1≤i<j≤4
P (Ei Ej Ek ) =
13
5
52 ,
5
1≤i<j<k≤4
P (E1 E2 E3 E4 ) = 0
Hence, by inclusion/exclusion
P
∪4i=1 Ei
4
=
1
39
5
52
5
4
−
2
26
5
52
5
4
+
3
13
5
52
5
=
6133
8330
8. If there are 12 strangers in a room, what is the probability that no two of them celebrate their birthday in
the same month? Clearly state the basic assumptions you are making.
Solution:
Assume that each person is equally likely to be born in any of the twelve months. Then the probability
that none of 12 individuals share the same birth-month is 12!/1212 ≈ 5.3723 × 10−5 .
Page 4 of 5
Math 421
Test 1
Dr. Pendergrass
9. Reorder the digits 1 through 9 randomly. Call a reordering deranged if digit d does not appear in place
d, for all d = 1, 2, · · · , 9. (So for instance, 825146973 is not deranged, since the digit 2 appears in place 2;
but the reordering 715832946 is deranged.) Find the probability that a reordering is deranged.
Solution:
This is the Matching Problem with N = 9. A deranged reordering corresponds to the situation in the
Matching Problem in which no man gets his own hat. The probability of this is
9
X
(−1)n
n=0
1
16687
=
≈ 0.3679
n!
45360
10. A man has n keys, one of which will open his door.
(a) If he tries the keys at random, discarding those that do not work, what is the probability that he will
open the door on his k th try?
(b) What is the probability if he does not discard previously tried keys?
Solution:
(a) Assume that all the n! orderings of the keys are equally likely. The door is opened on the k th try if
and only if the correct key is in position k in the reordering. The probability of this is
(n − 1)!
1
= ,
n!
n
k = 1, 2, 3, · · · , n
(b) If previous keys are not discarded, then on any try, the correct key is chosen with probability 1/n,
and an incorrect key is chosen with probability 1 − 1/n. It follows that the probability that the door
opens on the k th try is
1 k−1 1
1−
, , k = 1, 2, 3, · · ·
n
n
Page 5 of 5