1198/DCP1206 Probability, Fall 2015
5-Nov-2015
Homework 3 Solutions
Instructor: Prof. Wen-Guey Tzeng
Scribe: Amir Rezapour
1. From an urn that contains five red, five white, and five blue chips, we draw two chips
at random. For each blue chip we win $1, for each white chip we win $2, but for each
red chip we lose $3. If X represents the amount that we either win or we lose, what
are the possible values of X and probabilities associated with them?
Answer.
The set of possible values of X is {−6, −2, −1, 2, 3, 4}. The probabilities associated
with these values are
5
2
15 = 0.095,
2
5 5
1 1
= 15 = 0.238.
2
P (X = −6) = P (X = 2) = P (X = 4) =
P (X = −2) = P (X = −1) = P (X = 3)
(1)
2
2. From families with three children a family is chosen at random. Let X be the number
of girls in the family. Calculate and sketch the distribution function of X. Assume
that in a three-child family all gender distributions are equally probable.
Answer.
Let F be the distribution function of X. Then

0 t<0




 1/8 0 ≤ t < 1
1/2 1 ≤ t < 2
F (x) =


7/8 2 ≥ t < 3



1 t≥3
2
3. From 18 potential women jurors and 28 potential men jurors, a jury of 12 is chosen at
random. Let X be the number of women selected. Find the probability mass function
of X.
Answer.
Let p be the probability mass function of X; then
28 18
p(i) = P (X = i) =
i
12−i
46
12
i = 0, 1, 2, ..., 12.
(2)
2
1-1
4. From a drawer that contains 10 pair of gloves, six gloves are selected randomly. Let
X be the number of pairs of gloves obtained. Find the probability mass function of
X. Count the letter Y as a consonant.
Answer.
For i = 0, 1, 2, 3,we have
10
i
P (X = i) =
10−i
6−2i
6−2i .2
20
6
(3)
The numerical values of these probabilities are as follows.
i
p(i)
0
112/323
1
168/323
2
42/323
3
1/323
2
5. In a lottery, a player pays $1 and selects four distinct numbers from 0 to 9. Then,
from an urn containing 10 identical balls numbered from 0 to 9, four balls are drawn
at random and without replacement. If the numbers of three or all four of these balls
matches the players numbers, he wins $5 and $10, respectively. Otherwise, he loses.
On the average, how much money does the player gain per game? (Gain = win - loss.)
Answer.
Let X be the amount that the player gains in one game, then
4 6
1
3 1
P (X = 4) = 10 = 0.114, P (X = 9) = 10 = 0.005
4
(4)
4
and P (X = −1) = 1 − 0.114 − 0.005 = 0.881. Thus
E(X) = −1(0.881) + 4(0.114) + 9(0.005) = −0.38. Therefore, on the average, the
player loses 38 cents per game.
2
6. An ordinary deck of 52 cards is well-shuffled, and then the cards are turned face up
one by one until an ace appears. Find the expected number of cards that are face up.
Answer.
Let X be the number of cards to be turned face up until an ace appears. Let A be the
event that no ace appears among the first i − 1 cards that are turned face up. Let B
be the event that the ith card turned face up is an ace. We have
48
4
i−1
P (X = i) = P (AB) = P (B|A)P (A) =
. 52 (5)
52 − (i − 1) i−1
Therefore,
E(X) =
49
X
i
i=1
52
i−1
48
i−1
4
(53 − i)
To some, this answer might be counterintuitive.
1-2
= 10.6.
(6)
2
7. What are the expected number, the variance, and the standard deviation of the number of spades in a poker hand? (A poker hand is a set of five cards that are randomly
selected from an ordinary deck of 52 cards.)
Answer.
Clearly,
E(X) =
5
X
i
13
i
39
5−i
= 1.25.
52
5
i=0
39 5
X
i2 13
i
2
5−i = 2.426.
E(X ) =
52
5
i=0
Therefore, V ar(X) = 2.426 − (1.25)2 = 0.864, and hence σx =
(7)
√
0.864 = 0.9295.
2
8. A drunken man has n keys, one of which opens the door to his office. He tries the keys
at random, one by one, and independently. Compute the mean and the variance of the
number of trials required to open the door if the wrong keys (a) are not eliminated;
(b) are eliminated.
Answer.
(a) Let X be the number of trials required to open the door. Clearly,
P (X = x) = (1 −
Thus
E(x) =
∞
X
1 x−1 1
)
n, x = 1, 2, 3, ....
n
n
(8)
∞
x(1 −
1X
1
1 x−1 1
)
=
x(1 − )x−1
n
n
n
n
(9)
x=1
x=1
We know from calculus that ∀r, |r| < 1,
∞
X
xrx−1 =
x=1
Thus
∞
X
x(1 −
x=1
1
.
(1 − r)2
1 x−1
1
)
=
= n2 .
n
[1 − (1 − n1 )]2
(10)
(11)
Substituting (11) in (9), we obtain E(X) = n. To calculate V ar(X), first we find
E(X 2 ). We have
E(x2 ) =
∞
X
∞
x2 (1 −
x=1
1 x−1 1
1X 2
1
)
=
x (1 − )x−1
n
n
n
n
(12)
x=1
Now to calculate this sum, we multiply both sides of (10) by r and then differentiate
it with respect to r; we get
∞
X
x2 rx−1 =
x=1
1-3
1+r
.
(1 − r)3
(13)
Using this relation in (12), we obtain
E(x2 ) =
1 + 1 − n1
= 2n2 − n.
[1 − (1 − n1 )]3
(14)
Therefore,
V ar(X) = (2n2 − n) − n2 = n(n − 1).
(15)
(b) Let Ai be the event that on the ith trial the door opens. Let X be the number of
trials required to open the door. Then
P (X = 1) =
1
,
n
1 n−1
1
.
= ,
n−1 n
n
P (X = 3) = P (Ac1 Ac2 A3 ) = P (A3 |Ac2 Ac1 )P (Ac2 Ac1 )
P (X = 2) = P (Ac1 A2 ) = P (A2 |Ac1 )P (Ac1 ) =
(16)
= P (A3|Ac2 Ac1 )P (Ac2 |Ac1 )P (Ac1 )
1
1 n−2 n−1
.
.
=
=
n−2 n−1 n
n
Similarly, P (X = i) = 1/n for 1 ≤ i ≤ n. Therefore, X is a random number selected
from {1, 2, 3, ..., n}. E(X) = (n + 1)/2 and V ar(X) = (n2 − 1)/12.
2
1-4