M362K Homework Assignment 8 - solutions
Due October 21, 2011
Problems: 3.1.2, 3.1.4, 3.1.6, 3.1.8, 3.1.14, 4.5.2 from Pitman’s “Probability”.
Solution to 3.1.2:
a) Two draws are independent in this case, so P[X = k, Y = l] = P[X = k]P[Y = l]. The table looks
like this (last row represents the values of X and the first column those of Y ; the numbers in the
first row correspond to the (marginal) distribution of X; those in the last column to the (marginal)
distribution of Y ):
4
3
2
1
1
4
1
4
1
4
1
4
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
2
3
4
1
4
1
4
1
4
1
4
Y
X
The entries in the table that correspond to the event where X ≤ Y are those on upper-left half of
the table (including the diagonal). Therefore P[X ≤ Y ] = 85 .
b) When there is no replacement, all pairs with different components are equally likely, so the table is
given by
1
4
1
4
1
4
1
4
1
12
1
12
1
12
0
0
2
1
12
1
12
1
12
0
1
0
1
12
1
12
1
12
1
12
1
12
1
12
1
2
3
4
4
3
1
4
1
4
1
4
1
4
Y
X
The entries corresponding to X ≤ Y are the same ones as above, but now the probabilities sum to
P[X ≤ Y ] = 12 .
Solution to 3.1.4:
a) The table for (X1 , X2 ) is given by
6
5
4
3
2
1
1
6
1
6
1
6
1
6
1
6
1
6
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
36
1
2
3
4
5
6
1
6
1
6
1
6
1
6
1
6
1
6
X2
X1
b) The table for (Y1 , Y2 ) is given by
1
36
3
36
5
36
7
36
9
36
11
36
6
0
0
0
0
0
5
0
0
0
0
4
0
0
0
3
0
0
2
0
1
1
36
1
36
2
36
1
36
2
36
2
36
1
36
2
36
2
36
2
36
1
36
2
36
2
36
2
36
2
36
1
36
2
36
2
36
2
36
2
36
2
36
1
2
3
4
5
6
1
36
3
36
5
36
7
36
9
36
11
36
Y2
Y1
Solution to 3.1.6:
a) Clearly, supports of both X and Y are {0, 1, 2}. For a pair (k, l) with k, l ∈ {0, 1, 2}, we need to
count the number of possible outcomes of three coin tosses (such as HHT or THT) with exactly k
Hs among the first two tosses and l among the second two, and multiply by 18 . The table looks like
this
1
4
1
2
1
4
2
0
1
1
8
1
8
1
8
1
4
1
8
1
8
1
8
0
0
1
2
0
Y
X
2
1
4
1
2
1
4
b) They are not independent since, for example,
P[X = 0, Y = 0] =
1
8
6=
1
16
= P[X = 0] × P[Y = 0].
c) The support, i.e., the range, of X + Y is {0, 1, 2, 3, 4} its distribution is given by
k
0
1
2
3
4
P[X + Y = k]
1
8
2
8
2
8
2
8
1
8
You get the individual probability P[X + Y = k] by listing all the possible pairs that give you k as
the sum. For example
P[X + Y = 2] = P[X = 0, Y = 2] + P[X = 1, Y = 1] + P[X = 2, Y = 0] = 0 +
2
8
+ 0 = 28 .
Solution to 3.1.14:
a) Team A wins in g = 4 games with probability p4 (they have to win 4 games and loose none). For
g = 5, the team A have to win 4 out of the first 5 games, but not the first 4. The probability here
is 4p4 q. For the general g = 4, . . . , 7, the team A needs to win any 3 games
out of the first g − 1
g−1
(and lose the rest) and then win the last game. This can be done in 3 ways, and each one has
the probability p4 q g−4 , so the required probability is
g − 1 4 g−4
p q .
3
P
P
b) P[ team A wins ] = 7g=4 P[ A needs exactly g games to win ] = 7g=4
is no nicer expression.)
g−1
3
p4 q g−4 . (Note: There
c) For p = 2/3, the probability is 1808/2187 ∼
= 0.83
d) The point is that the teams could continue playing even after one of them wins 4 times. The losing
team could win at most 3 games so the probability of being the first who wins 4 games is the same
as the probability of winning at least 4 four games out of 7. (Note: the phrase at least is imporant.
The probability of winning exactly 4 games out of 7 is not the same as the probability of being the
first to win.)
e) Let G be the number of games until the first win. Then,
P[G = g] = P[ A wins in exactly g games ] + P[ B wins in exactly g games ]
g − 1 4 g−4
g − 1 4 g−4
=
p q
+
q p .
3
3
This expression will not depend on p only if p = 12 .
3
Solution to 4.5.2 (a):
1.0
0.8
0.6
0.4
0.2
-1
1
2
The value of FX is equal to 0 for x < 0 and equal to 1 for x ≥ 3.
4
3
4