Math/Stat 394 Homework 5
1. If we select two black balls then X = 4. This happens with probability
(42)
. If we select two white balls then X = −2. This happens with
(14
2)
(82)
. If we select two orange balls then X = 0. This
probability 14
(2)
(22)
happens with probability 14
. If we select one black and one white
(2)
ball then X = 1. This happens with probability 2·4·8
. If we select one
(12
2)
black and one orange ball then X = 2. This happens with probability
2·4·2
. If we select one white and one orange ball then X = −1. This
(12
2)
happens with probability 2·8·2
.
(12
2)
4. Five men and five women are ranked according to their scores on
an examination. Assume no two scores are alike and all 10! possible
rankings are equally likely. Let X denote the highest ranking achieved
by a women (for instance, X = 1 if the top-ranked person is female.)
Find P ({X = i}) for i = 1, 2, . . . , 10.
If the highest ranking woman has rank i then there are four women in
the bottom 10 − i. Thus
10−i
P (X = i) =
4
10
5
for i = 1, . . . , 6.
5. Let X denote the difference between the number of heads and the
number of tails obtained when a coin is tossed n times. What are the
possible values of X?
Let H be the number of heads and T be the number of tails. Then
T = n − H and X = H − T = 2H − n. As H can take on all integer
values between 0 and n X can take on all integer values between −n
and n which are of the same parity as n.
1
6. In problem 5, if the coin is assumed fair, for n = 3 what are the
probabilities associated with the values that X can take on?
P (X = −3) = P (X = 3) =
1
8
and
3
P (X = 1) = P (X = −1) = .
8
11. Admittedly sketchy proof.
(a) 333/1000, 200/1000, 142/1000, 66/1000, 9/1000. The limit as k
get large is 1/3, 1/5, 1/7, 1/15 and 1/105.
(b) Let
pk,n =
|{j ∈ 1, . . . , n: j has no repeated prime factor of p1 , . . . , pk }|
n
By the same reason as above
pk = lim pk,n →
k
Y
p2 − 1
n→∞
i
i=1
p2i
Taking the limit as k goes to we get
|{j ∈ 1, . . . , n: j has no repeated prime factor}|
n
→ lim pk
lim
n→∞
k
=
6
.
π2
13. A salesman has scheduled two appointments to sell encyclopedias. His
first appointment will lead to a sale with probability.3, and his second
will lead independently to a sale with probability .6. Any sale made is
equally likley to be either the deluxe model, which costs $1000 or the
standard model which cosets $500. Determine the probability mass
function of X, the total dollar value of all sales.
There possible outcomes are no sale (X = 0), one deluxe (X = 1000),
one regular (X = 500), two deluxe (X = 2000) one regular and one
deluxe (X = 1500) or two regular(X = 1000). No sales happen with
probability (.7)(.4) = .28. One sale happens with probability .3(.4) +
(.7)(.6) = .54. Thus one deluxe and one regular each happen with
probability .27. Two sales happens with probability (.3)(.6) = .18.
2
So two deluxe and two regular happen with probability .045 and one
regular and one deluxe happen with probability .09. Putting all this
together we get P (X = 0) = .28, P (X = 500) = .27, P (X = 1000) =
.27 + .045 = .315, P (X = 1500) = .09 and P (X = 2000) = .045.
14. Player one wins no rounds if her card is lower than the card for player
two. As all combinations are equally likely P (X = 0) = 1/2. Player
one wins one round if her card is higher than the card for player two
and lower than the card for player three. All six rankings of the first
three players are equally likely so P (X = 1) = 1/6. Thus P (X = 1) =
1/6. Player one wins one round if her card is higher than the card for
player two and three and lower than the card for player four. All twenty
four rankings of the first four players are equally likely and two of them
satisfy this condition (4123 and 4132). Thus P (X = 2) = 1/12. Player
one wins three rounds if she has the second highest card and player
4 has the highest card. The probability that player 4 has the highest
card is 1/5 and the conditional probability given that that player 1
has the next highest card is 1/4 thus P (X = 3) = 1/20. Player one
wins four rounds if she has the highest card. As all players are equally
likely to have the highest card P (X = 4) = 1/5
15. Let team i be the team with the ith worst record. They get 12 − i
balls in the lottery. Using notation from the next problem let Y1 = i
if team i gets the first pick, Y2 = i if team i gets the second pick, and
Y3 = i if team i gets the third pick.
P (X = 1) =
11
.
66
By Bayes’ rule the probability that team 1 gets the second pick is
P (X = 2) =
=
=
11
X
i=2
11
X
i=2
11
X
i=2
P (X = 2 and Y1 = i)
P (X = 2 | Y1 = i)P (Y1 = i)
12 − i
11
·
.
66
66 − (12 − i)
3
X
P (X = 3) =
P (X = 3, Y1 = i and Y2 = j)
i,j∈{2,...,11},i6=j
X
=
P (Y1 = i)P (Y2 = j | Y1 = i)P (X = 3 | Y1 = i and Y2 = j)
i,j∈{2,...,11},i6=j
X
=
i,j∈{2,...,11},i6=j
12 − i
12 − j
11
·
·
.
66
66 − (12 − i) 66 − (12 − i) − (12 − j)
X
P (X = 4) =
P (Y1 = i, Y2 = j and Y3 = k)
i,j,k∈{2,...,11},i6=j6=k,i6=k
X
=
P (Y1 = i)P (Y2 = j | Y1 = i)
i,j,k∈{2,...,11},i6=j6=k,i6=k
X
=
i,j,k∈{2,...,11},i6=j6=k,i6=k
16.
P (Y3 = k | Y1 = i and Y2 = j)
12 − j
12 − k
12 − i
·
·
.
66
66 − (12 − i) 66 − (12 − i) − (12 − j)
12 − m
.
66
By Bayes’ rule the probability that team m gets the second pick is
X
P (Y2 = m) =
P (Y2 = m and Y1 = i)
P (Y1 = m) =
i∈{1,...,11},i6=m
=
X
P (X = 2 | Y1 = i)P (Y1 = i)
i∈{1,...,11},i6=m
=
X
i∈{1,...,11},i6=m
P (Y3 = m) =
12 − m
12 − i
·
.
66
66 − (12 − i)
X
P (Y3 = m, Y1 = i and Y2 = j)
i,j∈{1,...,11},i6=j6=m,i6=m
=
X
P (Y1 = i)P (Y2 = j | Y1 = i)
i,j∈{1,...,11},i6=j6=m,i6=m
=
X
i,j∈{1,...,11},i6=j6=m,i6=m
4
·P (Y3 = m— Y1 = i and Y2 = j)
12 − i
12 − j
12 − m
·
·
.
66
66 − (12 − i) 66 − (12 − i) − (12 − j)
17. (a) We use the fact that
P (X = i) = lim P (X ≤ i)−P (X ≤ i−) = lim F (i)−F (i−) = F (i)−F (i− ).
→0
→0
Then we get P (X = 1) = F (1) − F (1− ) = 1/2 − 1/4 = 1/4.
P (X = 2) = F (2) − F (2− ) = 11/12 − 3/4 = 1/6. P (X = 3) =
F (3) − F (3− ) = 1 − 11/12 = 1/12.
(b) We use the fact that
P (X ∈ (i, j)) = lim P (X ≤ j−)−P (X ≤ i) = lim F (j−)−F (i) = F (j−)−F (i)
→0
→0
So P (X ∈ (.5, 1.5)) = 5/8 − 1/4 = 3/8.
5