# Math 461 Section P1 Quiz 3 Solution

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```Math 461 Section P1
Quiz 3 Solution
Feb 6, 2013
1. Sixty percent of students at a certain school wear neither a ring nor a
necklace. Twenty percent wear a ring and 30 percent wear a necklace.
If one of the students is chosen randomly, what is the probability that
this student is wearing:
(a) a ring or a necklace?
(b) a ring and a necklace?
Solution: If R (N , resp.) is the event of wearing a ring (necklace, resp.),
then P (R) = 20%, P (N ) = 30% and P ((R ∪ N )c ) = 60%; hence
(a) the event of wearing “either a ring or a necklace” is R ∪ N . The
probability of it is
P (R ∪ N ) = 1 − P ((R ∪ N )c ) = 1 − 60% = 40%.
(b) the event of wearing “a ring and a necklace” is R ∩ N . So the
probability is
P (R∩N ) = P (R)+P (N )−P (R∪N ) = 20%+30%−40% = 10%.
2. The game of craps is played as follows: a player rolls two dice. If the
sum is 2, 3, or 12, the player loses; if the sum is either a 7 or an 11,
the player wins. If the outcome is anything else, the player continues
to roll the dice until he rolls either the initial outcome or a 7. If the
7 comes first the player loses, whereas if the initial outcome reoccurs
before the 7 appears, the player wins. Compute the probability of a
player winning at craps.
Solution: If Ei is the event that the initial
P12outcome is i and the player wins,
then the probability of winning is i=2 P (Ei ) since Ei ’s are mutually
exclusive.
To compute P (Ei ), we first consider the experiment of rolling two dices
once. The sample space is S = {(i, j) | 1 ≤ i, j ≤ 6} where every pair
(i, j) occurs equally likely. The following are the outcomes corresponding to sums of two dices:
2:
3:
4:
5:
6:
7:
8:
9:
10 :
11 :
12 :
(1, 1);
(1, 2), (2, 1);
(1, 3), (2, 2), (3, 1);
(1, 4), (2, 3), (3, 2), (4, 1);
(1, 5), (2, 4), (3, 3), (4, 2), (5, 1);
(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1);
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2);
(3, 6), (4, 5), (5, 4), (6, 3);
(4, 6), (5, 5), (6, 4);
(5, 6), (6, 5);
(6, 6).
If Pi is the number of outcomes summing to i, then P2 = P12 = 1,
P3 = P11 = 2, P4 = P10 = 3, P5 = P9 = 4, P6 = P8 = 5 and P7 = 6 by
counting the outcomes.
(1 pt)
Now we consider two cases.
If i ∈ {2, 3, 7, 11, 12}, we know by the rule of game that P (Ei ) = 0 for
i = 2, 3, 12 and P (Ei ) = Pi /36 for i = 7, 11.
(1 pt)
If i ∈ {4, 5, 6, 8, 9, 10}, we define Ei,n to be the event of winning after
n rolls when the first roll has sum i. This is an experiment with 36n
equally likely outcomes. The event Ei,n happens when the first rolls
and the last roll get any of Pi outcomes summing to i, and the other
n − 2 rolls in between get any sum besides 7 and i, which have 36 −
Pi − P7 = 30 − Pi choices. By the basic principle of counting, Ei,n
contains Pi (30 − Pi )n−2 Pi = Pi2 (30 − Pi ) distinct outcomes, hence the
probability
P (Ei,n ) = Pi2 (30 − Pi )n−2 /36n .
(1 pt)
Since Ei,n (n = 2, 3, . . . ) are mutually exclusive,
P (Ei ) = P (
∞
[
Ei,n )
n=2
=
∞
X
P (Ei,n )
n=2
∞
X
Pi2 (30 − Pi )n−2
36n
n=2
k
∞ Pi2 X 30 − Pi
= 2
36 k=0
36
=
1
Pi2
2
36 1 − (30 − Pi )/36
Pi2
=
36(6 + Pi )
=
(1 pt)
for i ∈ {4, 5, 6, 8, 9, 10}.
Now by plugging in the values of Pi , we get the probability of a player
winning at craps is
2
1
1
1
3
42
52
244
+
+2·
+
+
=
≈ 0.493. (1 pt)
6 18
36 6 + 3 6 + 4 6 + 5
295
```