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McGill University
Probability Theory (MATH323B)
Mid-Term Exam, Tuesday March 3, 2009
Instructions: This test consists of four main and one bonus questions.
There is no optional question among the main questions. Please answer all
four of them.
Question 1. A company orders supplies from M distributors and wishes
to place n orders (n < M ). Assume that the company places the orders in a
manner that allows every distributors an equal chance of obtaining any one
order and there is no restriction on the number of orders that can be placed
with any distributor. Find the probability that a particular distributor, say
distributor I, gets exactly k orders (k ≤ n). (6 marks)
Solution:
First Approach: The number of ways that k out of n orders can be set
aside for distributor I is nk . The rest of the orders, i.e. (n − k), can be
freely distributed among the rest of the distributors, i.e. (M − 1). We have
(M − 1)n−k ways to distributed the n − k orders among (M − 1) distributors.
The total number of ways we can distribute n orders among M distributors
is M n . Let A = {Distributor I gets exactly k orders}. Thus using mn rule
we have
k n−k
n
(M − 1)n−k
n
1
1
k
=
P (A) =
1−
.
k
Mn
M
M
Second Approach: Define
(
1,
Xi =
0,
if distributor I gets the i-th order,
otherwise.
Note that X
are independent and P (Xi = 1) = 1/M for i = 1, 2, · · · , n.
Pi ’s
n
Thus Y = i=1 Xi is the number of orders that distributor I gets. We now
note that Y ∼ Bin(n, 1/M) and
k n−k
n
1
1
P (A) = P (Y = k) =
1−
.
k
M
M
1
Question 2. As items come to the end of a production line, an inspector
chooses which items are to go through a complete inspection. Ten percent
of all items produced are defective. Sixty percent of all defective items go
through a complete inspection, and 20% of all good items go through a
complete inspection. Given that an item is completely inspected, what is the
probability it is defective? (6 marks)
Solution:
Define D = {An item chosen randomly from the production line being defective}
and I = {An item goes through complete inspection}. We then have P (D) =
0.1, P (I | D) = 0.6 and P (I | Dc ) = 0.2. We need to find P (D | I).
P (I | D)P (D)
P (I | D)P (D) + P (I | Dc )P (Dc )
1
(0.6) × (0.1)
= = 25%
=
(0.6) × (0.1) + (0.2) × (0.9)
4
P (D | I) =
2
Question 3. The telephone lines serving an airline reservation office are
all busy about 60%of the time.
(a) If you are calling this office, what is the probability that you will complete
your call on the first try? The second try? The third try? (4 marks)
Solution:
Let X be the number of tries to the first success. Then X ∼ Geometric(p =
0.4) and hence
P (X = k) = p(1 − p)k−1 = 0.4 × 0.6k−1 ,
for k = 1, 2, 3, ...
In particular, P (X = 1) = 0.4, P (X = 2) = 0.24, and P (X = 3) = 0.144.
(b) If you and a friend must both complete calls to this office, what is the
probability that a total of four tries will be necessary for both of you to get
through? (3 marks)
Solution:
Let Y be the number of success to the second success. Then Y ∼
N Bin(r = 2, p = 0.4) and hence
k−1 r
P (Y = k) =
p (1−p)k−r = (k−1)×0.42 ×0.6k−2 , for k = 2, 3, 4, ...
r−1
In particular, P (Y = 4) = 3 × 0.42 × 0.62 = 0.1728.
3
Question 4. The number of bacteria colonies of a certain type in samples
of polluted water has a Poisson distribution with a mean of 2 per cubic
centimeter.
(a) If four 1-cubic-centimeter samples are independently selected from this
water, find the probability that at least one sample will contain one or more
bacteria colonies.(3 marks)
Solution:
Let W denotes the number of bacteria colonies in a sample of 1-cubiccentimeter. Then W ∼ P o(λ = 2). Define, for i = 1, 2, 3, 4,
(
1, if the i-th sample contains one or more bacteria colonies
Xi =
0, otherwise.
Note that
e−2 20
= 1 − e−2 .
P (Xi = 1) = P (W ≥ 1) = 1 − P (W = 0) = 1 −
0!
P4
and Y = i=1 Xi ∼ Bin(n = 4, p = 1 − e−2 ) where Y count the number of
samples containing one or more bacteria. We need then find
P (Y ≥ 1) = 1 − P (Y = 0) = 1 − (1 − p)4 = 1 − e−8 .
(b) How many 1-cubic-centimeter samples should be selected in order to have
a probability of approximately 0.95 of seeing at least one bacteria colony? (3
marks)
Solution:
Following the argument in part (a), to find the required sample size n∗
we should solve the following equation
P (Y
∗
1) = 1 − e−2n = 0.95.
It is easy to see that n∗ = − ln(0.05)/2.
4
Bonus Question. Two gamblers bet $1 each on the successive tosses of
a coin. Each has a bank of $6.
(a) What is the probability that they break even after six tosses of the coin?
(3 marks)
Solution:
They can break even if and only if we have three ”H” and three ”T”.
Probability of such even is therefore
6
5
6
1
= .
2
16
3
(b) What is the probability that one player, say, Jones, wins all the money
on the tenth toss of the coin? (7 marks)
Solution:
This event can only happen if Jones wins 8 times, 8 ”H”, and loses 2
times, 2 ”T”. Now to make sure that the game is not terminated at the 6th
or the 8th trial, note that the game can only be terminated at even number
of trials, we should have at least one ”T” among the first 6 trials and at
least one ”T” between the 7th and 8th trials. The probability of interest is
therefore
10
27
6 2
6
1
=
.
+
2
1024
1 1
2
A more systematic approach: Random Walk
This problem can be systematically studied using Random Walks. Suppose both players start with m dollars in their pockets. Define, for i =
1, 2, 3, ..., n,
(
1, if Jones wins the i-th trial,
Wi =
−1, otherwise,
and Xi = (1/2)Wi + (1/2). Note that Wi ’s are independent and hence so
are Xi ’s. Note also that Xi ’s are Bernoulli variable whose probability
of
Pn
success does not change from one trial P
to another and thus Yn = i=1 Xi ∼
Bin(n, 1/2). On the other hand, Vn = ni=1 Wi = 2Yn − n. The probability
of interest is
P (Vn = 2m | Vk < 2m ∀k < n).
Using the connection made above between Vn and Yn ∼ Bin(n, 1/2) we can
easily find such probabilities.
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