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1206/DCP1206 Probability, Fall 2014
17-Nov-2014
Homework 3 Solutions
Instructor: Prof. Wen-Guey Tzeng
1. In a society of population N, the probability is p that a person has a certain rare
disease independently of others. Let X be the number of people who should be tested
until a person with the disease is found, X = 0 if no one with the disease is found.
What are the possible values of X? Determine the probabilities associated with these
values.
Answer. The set of possible values of X is {0, 1, 2, . . . , N }. Assuming that people
have the disease independent of each other,
(1 − p)i−1 p 1 ≤ i ≤ N
(1)
P (X = i) =
(i − p)N
i=0
2
2. A grocery store sells X hundred kilograms of rice every day, where the distribution of
the random variable X is of the following form:

0
x<0



kx2
0≤x<3
F (x) =
(2)
k(−x2 + 12x − 3) 3 ≤ x < 6



1
x≥6
Suppose that this grocery stores total sales of rice do not reach 600 kilograms on any
given day.
Answer. Note that X is neither continuous nor discrete.
(a) Find the value of k.
F (6−) = 1 implies that k(−36 + 72 − 3) = 1; so k = 1/33.
(b) What is the probability that the store sells between 200 and 400 kilograms of
rice next Thursday?
F (4) − F (2) = 29/33 − 4/33 = 25/33.
(c) What is the probability that the store sells over 300 kilograms of rice next Thursday?
1 − F (3) = 1 − (24/33) = 9/33.
(d) We are given that the store sold at least 300 kilograms of rice last Friday. What
is the probability that it did not sell more than 400 kilograms on that day?
(3−)
29/33−9/33
P (X ≤ 4|X ≥ 3) = F (4)−F
= 56 .
1−F (3−) =
1−9/33
2
1-1
3. In a small town there are 40 taxis, numbered 1 to 40. Three taxis arrive at random at
a station to pick up passengers. What is the probability that the number of at least
one of the taxis is less than 5?
Answer. Let X be the minimum of the three numbers,
(36
3)
P (X < 5) = 1 − P (X ≥ 5) = 1 − 40
= 0.277
2
(3)
4. In the experiment of rolling a balanced die twice, let X be the sum of the two numbers
obtained. Determine the probability mass function of X.
Answer. The possible values of X are 2, 3, ... , 12. The sample space of this
experiment consists of 36 equally likely outcomes. Hence the probability of any of
them is 1/36. Thus
p(2) = P (X = 2) = P ({(1, 1)}) = 1/36,
p(3) = P (X = 3) = P ({(1, 2), (2, 1)}) = 2/36,
p(4) = P (X = 4) = P ({(1, 3), (2, 2), (3, 1)}) = 3/36.
Similarly,
p(5) = 4/36, p(6) = 5/36, p(7) = 6/36, p(8) = 5/36, p(9) = 3/36, p(10)2/36, p(12) =
1/36.
2
5. A binary digit or bit is a zero or one. A computer assembly language can generate
independent random bits. Let X be the number of independent random bits to be
generated until both 0 and 1 are obtained. Find the probability mass function of X
Answer. The set of possible values of X is {2, 3, 4, . . .}. For n ≥ 2, X = n if and
only if either all of the first n1 bits generated are 0 and the nth bit generated is 1, or
all of the first n1 bits generated are 1 and the nth bit generated is 0. Therefore, by
independence,
p(X = n) = (1/2)n−1 .(1/2) + (1/2)n−1 .(1/2) = (1/2)n−1 , n ≥ 2.
2
6. A fair die is tossed successively. Let X denote the number of tosses until each of the
six possible outcomes occurs at least once. Find the probability mass function of X.
Hint: For 1 ≤ i ≤ 6, let Ei be the event that the outcome i does not occur during
the first n tosses of the die. First calculate P (X > n) by writing the event X > n in
terms of E1 , E2 , . . . , E6 .
S
Answer. Clearly, p(X > n) = P ( 6i=1 Ei ).
To calculate P (E1 ∪ E2 ∪ . . . ∪ E6 ), we use the inclusion-exclusion principle. To do
so, we must calculate the probabilities of all possible intersections of the events from
E1 , . . . , E6 , add the probabilities that are obtained by intersecting an odd number
of events, and subtract all the probabilities
that are obtained by intersecting
an even
6
6
number of events. Clearly, there are 1 terms of the form P (Ei ), 1 terms of the form
P (Ei Ej ), 61 terms of the form P (Ei Ej Ek ), and so on. Now for all i, P (Ei ) = (5/6)n ;
for all i and j, P (EiEj) = (4/6)n ; for all i, j, and k, P (Ei Ej Ek ) = (3/6)n ; and so on.
Thus
P (X > n) = P (E1 ∪ E2 ∪ . . . ∪ E6 ) = 61 (5/6)n − 62 (4/6)n + 63 (3/6)n − 64 (2/6)n +
6
n
n
n
n
n
n
5 (1/6) = 6(5/6) − 15(4/6) + 20(3/6) − 15(2/6) + 6(1/6) .
1-2
Let p be the probability mass function of X. The set of all possible values of X is
{6, 7, 8, . . .}, and
p(n) = P (X = n) = P (X > n − 1) − p(X > n) = (5/6)n−1 − 5(4/6)n−1 + 10(3/6)n−1 −
10(2/6)n−1 + 5(1/6)n−1 , n ≥ 6
2
7. In a lottery every week, 2,000,000 tickets are sold for $1 apiece. If 4000 of these tickets
pay off $30 each, 500 pay off $800 each, one ticket pays off $1,200,000, and no ticket
pays off more than one prize, what is the expected value of the winning amount for a
player with a single ticket?
Answer. The expected value of the winning amount is
4000
500
1
30 2000000
+ 800 2000000
+ 1200000 2000000
= 0.86
Considering the cost of the ticket, the expected value of the players gain in one game
is −1 + 0.86 = −0.14.
2
8. The distribution function of a random variable X is given by

0 x < −3




 3/8 −3 ≤ x < 0
1/2 0 ≤ x < 3
F (x) =


 3/4 3 ≤ x < 4


1 x≥4
(3)
Calculate E(X), E(X 2 − 2|X|), and E(X|X|).
Answer. p(x) the probability mass function of X is given by
x
p(x)
-3
3/8
0
1/8
3
1/4
4
1/4
Hence
E(X) = −3.3/8 + 0.1/8 + 3.1/4 + 4.1/4 = 5/8,
E(X 2 ) = 9.3/8 + 0.1/8 + 9.1/4 + 16.1/4 = 77/8,
E(|X|) = 3.3/8 + 0.1/8 + 3.1/4 + 4.1/4 = 23/8,
E(X 2 − 2|X|) = 77/8 − 2(23/8) = 31/8,
E(X|X|) = −9.3/8 + 0.1/8 + 9.1/4 + 16.1/4 = 23/8.
2
9. Suppose that there exist N families on the earth and that the maximum number of
children a family
has is c. For j = 0, 1, 2, ..., c , let α be the fraction of families with
P
j children ( cj=0 αj = 1) . A child is selected at random from the set of all children
in the world. Let this child be the Kth born of his or her family; then K is a random
variable. Find E(K).
Answer. Let Aj be the event that the person belongs to a family with j children.
Then
P
P
P (K = k) = cj=0 P (K = k|Ai )P (Ai ) = cj=k 1j .αj
Therefore,
P
P
P
P
P
E(K) = ck=1 kP (K = k) = ck=1 k cj=1 1j .αj = ck=1 cj=1 kj .αj
2
1-3
10. Mr. Jones is about to purchase a business. There are two businesses available. The
first has a daily expected profit of $150 with standard deviation $30, and the second
has a daily expected profit of $150 with standard deviation $55. If Mr. Jones is
interested in a business with a steady income, which should he choose?
Answer. On average, in the long run, the two businesses have the same profit. The
one that has a profit with lower standard deviation should be chosen by Mr. Jones
because hes interested in steady income. Therefore, he should choose the first business.
2
11. Find the variance of X, the random variable with probability mass function
(|x − 3| + 1)/28 x = −3, −2, −1, 0, 1, 2, 3
p(x) =
0
otherwise.
(4)
P3
P3
2
2
Answer. E(X) =
x=−3 xp(x) = −1, E(X ) =
x=−3 x p(x) = 4. Therefore,
V ar(X) = 4 − 1 = 3.
2
12. Suppose that X is a discrete random variable with E(X) = 1 and E[X(X − 2)] = 3.
Find V ar(−3X + 5).
Answer. By the Corollary of Theorem 4.2, E(X 2 − 2X) = 3 implies that E(X 2 ) −
2E(X) = 3. Substituting E(X) = 1 in this relation gives E(X 2 ) = 5. Hence, by
Theorem 4.3, V ar(X) = E(X 2 ) − [E(X)]2 = 5 − 1 = 4.
By Theorem 4.5,
V ar(−3X + 5) = 9V ar(X) = 9 × 4 = 36.
2
1-4
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