Chapter #4 – Discrete Random Variables
Question #1: Two balls are chosen randomly from an urn with 8 white, 4 black and 2 orange
balls. Suppose that we win $2 for each black ball selected and we lose $1 for each white ball
selected (nothing happens with the orange balls). Let X denote our winnings. What are the
possible values of X, and what are the probabilities associated with each value?
 There are 14 total balls in the urn. The possible values are 𝑋 = {−2, −1,0,1,2,4}, which
are computed from the six unique combinations of balls that can be drawn. For
example, two white balls will yield –$2 while one black and one white ball will yield
winnings of $1. The following are the desired probabilities.
o 𝑝(𝑋 = −2) =
o 𝑝(𝑋 = −1) =
𝐶(8,2)
𝐶(14,2)
𝐶(2,1)∗𝐶(8,1)
𝐶(14,2)
𝐶(2,2)
o 𝑝(𝑋 = 0) = 𝐶(14,2)
o 𝑝(𝑋 = 1) =
o 𝑝(𝑋 = 2) =
𝐶(8,1)∗𝐶(4,1)
𝐶(14,2)
𝐶(2,1)∗𝐶(4,1)
𝐶(14,2)
𝐶(4,2)
o 𝑝(𝑋 = 4) = 𝐶(14,2)
(white + white)
(orange + white)
(orange + orange)
(white + black)
(orange + black)
(black + black)
Question #5: Let X represent 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 𝑖 denotes the number of heads and 𝑗 denotes the number of tails when a coin is
tossed 𝑛 times. Since 𝑖 + 𝑗 = 𝑛, we have 𝑗 = 𝑛 − 𝑖 or 𝑖 = 𝑛 − 𝑗. Therefore, we can see
that 𝑋 = 𝑖 − 𝑗 = 𝑖 − (𝑛 − 𝑖) = 2𝑖 − 𝑛 for 𝑖 = 0,1, … , 𝑛. In other words, the possible
values for the random variable are 𝑋 = {−𝑛, 2 − 𝑛, 4 − 𝑛, … , 𝑛}.
Question #6: In Problem 5, for 𝑛 = 3, if the coin is assumed fair, what are the probabilities
associated with the values that X can take on?
 We see that 𝑋 = {−3, −1,1,3}. After listing the outcomes, the following are the desired
probabilities for the four values of the random variables.
o 𝑃(𝑋 = −3) = 1/8
o 𝑃(𝑋 = −1) = 3/8
o 𝑃(𝑋 = 1) = 3/8
o 𝑃(𝑋 = 3) = 1/8
Question #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 likely to be either for the deluxe model,
which costs $1000, or the standard model, which costs $500. Determine the probability mass
function of X, the total dollar value of all sales.
 If we let ∅ denote the event of no sale, 𝑆 denote a standard sale and 𝐷 denote a deluxe
sale, then the sample space 𝑆 = {∅∅, ∅𝑆, ∅𝐷, 𝑆𝑆, 𝑆∅, 𝑆𝐷, 𝐷𝐷, 𝐷∅, 𝐷𝑆} while the random
variable 𝑋 assigns the following real numbers to these outcomes:
o 𝑋(∅∅) =
0
o 𝑋(∅𝑆) = 𝑋(𝑆∅) =
500
o 𝑋(∅𝐷) = 𝑋(𝐷∅) = 𝑋(𝑆𝑆) =
1000
o 𝑋(𝑆𝐷) = 𝑋(𝐷𝑆) =
1500
o 𝑋(𝐷𝐷) =
2000

If we recall that 𝑃(𝑠𝑎𝑙𝑒 1) = 0.3, 𝑃(𝑠𝑎𝑙𝑒 2) = 0.6 and 𝑃(𝑆 𝑠𝑎𝑙𝑒) = 𝑃(𝐷 𝑠𝑎𝑙𝑒) = 0.5,
the probabilities to include in the probability mass function are calculated as follows.
o 𝑃(𝑋 = 0) = (1 − 0.3)(1 − 0.6) = (0.7)(0.4)
o 𝑃(𝑋 = 500) = (0.7)(0.6)(0.5) + (0.3)(0.5)(0.4)
o 𝑃(𝑋 = 1000) = (0.7)(0.6)(0.5) + (0.3)(0.5)(0.4) + (0.3)(0.5)(0.6)(0.5)
o 𝑃(𝑋 = 1500) = (0.3)(0.5)(0.6)(0.5) + (0.3)(0.5)(0.6)(0.5)
o 𝑃(𝑋 = 2000) = (0.3)(0.5)(0.6)(0.5)
Question #21: Four buses carrying 148 students from the same school arrive at a football
stadium. The buses carry, respectively, 40, 33, 25, and 50 students. One of the students is
randomly selected. Let 𝑋 denote the number of students that were on the bus carrying the
randomly selected student. One of the 4 bus drivers is also randomly selected. Let 𝑌 denote
the number of students on her bus. (a) Which of 𝐸[𝑋] or 𝐸[𝑌] do you think is larger? Why?
(b) Compute 𝐸[𝑋] and 𝐸[𝑌] in this experiment.
a) In the first sampling method, we draw one of 148 students at random. Naturally, the
probability of choosing a student from the fullest bus is higher than the probability of
drawing a student from the other buses. Thus, the fuller buses are weighted higher
than the other buses, so we would expect 𝐸(𝑋) to be larger. In the second method,
one of the four bus drivers is randomly selected. Each bus driver is equally likely to
be chosen! Each bus is weighted the same in this came, therefore, 𝐸(𝑌) < 𝐸(𝑋).
Equality holds when each bus contains the same number of students.
40
33
25
50
b) 𝐸[𝑋] = 40 ∗ 148 + 33 ∗ 148 + 25 ∗ 148 + 50 ∗ 148 = 39.28
1
1
1
1
𝐸[𝑌] = 40 ∗ + 33 ∗ + 25 ∗ + 50 ∗ = 37
4
4
4
4
Question #22: Suppose that two teams play a series of games that ends when one of them
has won 𝑖 games. Suppose that each game played is, independently, won by team A with
probability 𝑝. Find the expected number of games that are played when we have (a) 𝑖 = 2
and (b) 𝑖 = 3. Also, show in both cases that this number is maximized when 𝑝 = 1/2.
a) Let 𝑋 denote the number of games played. When 𝑖 = 2, the random variable 𝑋 can be
either 2 (one teams wins both) or 3 (the teams split the first two games and winner
of the third wins the game). Therefore, we can compute the expected value of 𝑋 as
follows: 𝐸[𝑋] = 2[(2−1
)𝑝2 + (2−1
)(1 − 𝑝)2 ] + 3[(3−1
)𝑝2 (1 − 𝑝) + (3−1
)𝑝(1 − 𝑝)2 ] =
2−1
2−1
2−1
2−1
−2𝑝2 + 2𝑝 + 2. To find where the expected value is maximized, we differentiate and
𝑑
1
set equal to zero to solve for the probability 𝑝: 𝑑𝑝 𝐸[𝑋] = −4𝑝 + 2 = 0 → 𝑝 = 2 (this
𝑑2
is a maximum since the second derivative is negative, 𝑑𝑝2 𝐸[𝑋] = −4 < 0).
b) When 𝑖 = 3, the random variable 𝑋 can be either 3, 4 or 5. We thus calculate as follows
𝐸[𝑋] = 3[(3−1
)𝑝3 + (3−1
)(1 − 𝑝)3 ] + 4[(4−1
)𝑝3 (1 − 𝑝) + (4−1
)𝑝(1 − 𝑝)3 ] +
3−1
3−1
3−1
3−1
5[(5−1
)𝑝3 (1 − 𝑝)2 + (5−1
)𝑝2 (1 − 𝑝)3 ] = 6𝑝4 − 12𝑝3 + 3𝑝2 + 3𝑝 + 3. Then we see
3−1
3−1
that
𝑑
𝑑𝑝
1
𝐸[𝑋] = 24𝑝3 − 36𝑝2 + 6𝑝 + 3 = 0 → 𝑝 = 2.
Question #36: Consider Problem 22 above with 𝑖 = 2. Find the variance of the number of
games played, and show that this number is maximized when 𝑝 = 1/2.
 𝑉𝑎𝑟(𝑋) = 𝐸[𝑋 2 ] − [𝐸(𝑋)]2 = {22 [𝑝2 + (1 − 𝑝)2 ] + 32 [2𝑝2 (1 − 𝑝) + 2𝑝(1 − 𝑝)2 ]} −
{−2𝑝2 + 2𝑝 + 2}2 = −4𝑝4 + 8𝑝3 − 6𝑝2 + 2𝑝. We then calculate the derivative to
optimize:
𝑑
𝑑𝑝
1
𝑉𝑎𝑟(𝑋) = −16𝑝3 + 24𝑝2 − 12𝑝 + 2 = 0 → 𝑝 = .
2
Question #26: One of the numbers 1 through 10 is randomly chosen. You are to try to guess
the number chosen by asking questions with “yes–no” answers. Compute the expected
number of questions you will need to ask in each of the following two cases: (a) Your ith
question is to be “is it i?” for 𝑖 = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. (b) With each question you try to
eliminate one half of the remaining numbers, as nearly as possible.
1
a) Let 𝑄 be the number of questions asked. It then follows that 𝑃(𝑄 = 𝑖) = 10 for all 𝑖.
1
1
1
This, we have that 𝐸(𝑄) = 1 ∗ 10 + 2 ∗ 10 + ⋯ + 10 ∗ 10 = 5.5.
b) Consider the following strategy: divide what you have in halves and ask, “is it greater
than or equal to 𝑥?” where 𝑥 is the middle of the numbers. For instance, your first
question is, “is it greater than or equal to 5.5?”. (All other strategies of this type are
similar. But you need to be consistent.) Let 𝑋 denote the random number, so that
1
𝑃(𝑋 = 𝑖) = 10 for all 𝑖. If 𝑋 = {1,2,3,6,7,8}, then 𝑄 = 3 and if 𝑋 = {4,5,9,10}, then we
see that 𝑄 = 4. Therefore, we have the following results:
 𝑃(𝑄 = 3) = 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + ⋯ + 𝑃(𝑋 = 8) = 0.6
 𝑃(𝑄 = 4) = 𝑃(𝑋 = 4) + 𝑃(𝑋 = 5) + ⋯ + 𝑃(𝑋 = 10) = 0.4
 𝐸[𝑄] = 3(0.6) + 4(0.4) = 3.4.
Question #41: A man claims to have extrasensory perception. As a test, a fair coin is flipped
10 times and the man is asked to predict the outcome in advance. He gets 7 out of 10 correct.
What is the probability that he would have done at least this well if he had no ESP?

1
If we let 𝑋 be the number of correct guesses, then we see 𝑋~𝐵𝐼𝑁(10, 2). The
probability that he would have done at least as well as seven correct guess is thus
1 7 1 3
1 10 1 0
given by 𝑃(𝑋 ≥ 7) = (10
) (2) (2) + ⋯ + (10
) (2)
7
10
1 𝑖
1 10−𝑖
10
(2) = ∑10
𝑖=7( 𝑖 ) (2) (2)
.
Question #46: Suppose that it takes at least 9 votes from a 12-member jury to convict a
defendant. Suppose also that the probability that a juror votes a guilty person innocent is 0.2,
whereas the probability that the juror votes an innocent person guilty is 0.1. If each juror
acts independently and if 65% of the defendants are guilty, find the probability that the jury
renders a correct decision. What percentage of defendants is convicted?
 Let the events be 𝑉𝐼 (vote innocent), 𝑉𝐺 (vote guilty), 𝐺 (guilty) and 𝐼 (innocent), so
we are given 𝑃(𝑉𝐼 |𝐺) = 0.2, 𝑃(𝑉𝐺 |𝐼) = 0.1, 𝑃(𝐺) = 0.65 and 𝑃(𝐼) = 0.35. Thus, if 𝐶 is
conviction, what we want to calculate is 𝑃(𝐶) = 𝑃(𝐶|𝐺)𝑃(𝐺) + 𝑃(𝐶|𝐼)𝑃(𝐼), where
12
𝑖
12−𝑖
o 𝑃(𝐶|𝐺) = ∑12
≈ 0.79
𝑖=9( 𝑖 )(0.8) (0.2)
o 𝑃(𝐺) = 0.65
o 𝑃(𝐼) = 0.35
12
𝑖
12−𝑖
o 𝑃(𝐶|𝐼) = ∑12
≈ 0.00000017
𝑖=9( 𝑖 )(0.1) (0.9)

This, the answer is 𝑃(𝐶) = (0.79)(0.65) + (0.00000017)(0.35) ≈ 0.5165.
Question #56: How many people are needed so that the probability that at least one of them
has the same birthday as you is greater than 1/2?
 The number of people in a random collection of size 𝑛 that have the same birthday as
𝑛
yourself is approximately Poisson distributed with mean 𝜆 = 365. Hence, the
probability that at least one person has the same birthday as you is approximately
equal to 1 − 𝑃(𝑛𝑜𝑛𝑒) = 1 −
𝑒 −𝜆 𝜆0
0!
𝑛
𝑛
1
= 1 − 𝑒 −𝜆 = 1 − 𝑒 −365 . Since 𝑒 −𝑥 = 2 when we
1
𝑛
have 𝑥 = 𝑙𝑛(2), we have 1 − 𝑒 −365 ≥ 2 when 365 ≥ 𝑙𝑛(2) so 𝑛 ≥ 365 ∗ 𝑙𝑛(2) = 253.
Question #71: Consider a roulette wheel consisting of 38 numbers 1 through 36, 0 and
double 0. If Smith always bets that the outcome will be one of the numbers 1 through 12,
what is the probability that (a) Smith will lose his first 5 bets, (b) his first win will occur on
his fourth bet?
26 5
a) (38) = 0.15
26 3 12
b) (38) (38) = 0.10
Question #82: A purchaser of transistors buys them in lots of 20. It is his policy to randomly
inspect 4 components from a lot and to accept the lot only if all 4 are nondefective. If each
component in a lot is, independently, defective with probability 0.1, what proportion of lots
is rejected by the purchaser?
 𝑃(𝑟𝑒𝑗𝑒𝑐𝑡) = 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒) = 1 − 𝑃(𝑛𝑜 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒) = 1 − (0.9)4 ≈ 0.3.