CS 3341 HW # 3 SOLUTION

1. Let the random variable X denote the number of heads in three independent tosses of
a fair coin. Find the PMF of X. Also find the CDF of X.

PMF of X:

x p(x)
0 p(0) = P(X = 0) = P(TTT) = 1/8
1 p(1) = P(X = 1) = P(HTT) + P(THT) + P(TTH) = 3/8
2 p(2) = P(X = 2) = P(HHT) + P(HTH) + P(THH) = 3/8
3 p(3) = P(X = 3) = P(HHH) = 1/8

CDF of X:

F(0) = P(X ≤ 0) = P(0 or fewer heads) = 1/8

F(1) = P(X ≤ 1) = P(1 or fewer heads) = (1/8) + (3/8) = 4/8
F(2) = P(X ≤ 2) = P(2 or fewer heads) = (1/8) + (3/8) + (3/8) = 7/8
F(3) = P(X ≤ 3) = P(3 or fewer heads) = (1/8) + (3/8) + (3/8) + (1/8) = 1

Hence, the CDF of X is:

 0 x<0
1 / 8 0 ≤ x < 1

F ( x ) = 4 / 8 1 ≤ x < 2
7 / 8 2 ≤ x < 3

 1 3≤ x

2. Let X denote the number of busy servers at the checkout counters in a store at 5pm.
Suppose that the CDF of X is:
 0 x<0
0.20 0 ≤ x < 1

0.50 1 ≤ x < 2
F ( x) = 
0.80 2 ≤ x < 3
0.90 3 ≤ x < 4

 1 4≤ x

(a) Find the PMF of X.

The possible values of X are the points at which the CDF F(x) jumps, and the sizes of
the jumps are the probabilities of taking those values. Hence the PMF of X is:

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 x p(x)
0 p(0) = P(X = 0) = F(0) – F(0-)* = 0.20 – 0 = 0.20
1 p(1) = P(X = 1) = F(1) – F(1-) = 0.50 – 0.20 = 0.30
2 p(2) = P(X = 2) = F(2) – F(2-) = 0.80 – 0.50 = 0.30
3 p(3) = P(X = 3) = F(3) – F(3-) = 0.90 – 0.80 = 0.10
4 p(4) = P(X = 4) = F(4) – F(4-) = 1 – 0.90 = 0.10

*F(a-) represents P(X < a).

(b) Find P(X > 3).

P(X > 3) = 1 – P(X ≤ 3) = 1 – F(3) = 1 – 0.90 = 0.10

3. It is known that there is a defective chip on a computer board that contains 8 chips. A
technician tests the chips one at a time until the defective chip is found. Assume that
the chip to be tested is selected at random without replacement. Let the random
variable X denote the number of chips tested. Find the PMF of X.

Total of 8 chips; 7 = (G)ood, 1 = (D)efective. The PMF is:

x P(X = x)
1 P(X = 1) = P(D) = 1/8
2 P(X = 2) = P(GD) = (7/8)(1/7) = 1/8
3 P(X = 3) = P(GGD) = (7/8)(6/7)(1/6) = 1/8
4 P(X = 4) = P(GGGD) = (7/8)(6/7)(5/6)(1/5) = 1/8
5, 6, 7, 8 P(X = 5) = P(X = 6) = P(X = 7) = P(X = 8) =1/8 [as before]

4. Let X denote the number of speeding tickets issued by Officer Smith on Monday, and
let Y denote the number given on Tuesday. The PMF of X and Y are

X=k 0 1 2 3 4
P(X=k) 0.1 0.2 0.3 0.2 0.2

Y=m 0 1 2 3 4 5
P(Y=m) 0.1 0.1 0.2 0.3 0.2 0.1

Assume that X and Y are independent. Find the joint PMF of X and Y. Also find the
PMF of X + Y.

The table of joint PMF can be obtained by using P(X = k, Y = m) = P(X=k) P(Y=m),
since X and Y are independent.

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 The possible values of Z = X + Y are 0, 1, 2, …, 9. Here is the PMF of Z:

z 0 1 2 3 4 5 6 7 8 9
P(Z=z) 0.01 0.03 0.07 0.12 0.18 0.20 0.18 0.13 0.06 0.02

One example of how we get this table is the following:

P(Z = 1) = P[ (X, Y) = (0, 1) or (1, 0)] = P[(1,0)] + P[(0,1)], and now substitute these
probabilities from the joint PMF table.

5. The joint distribution of random variables X and Y is given in the table,

(a) Find the marginal PMF’s of X and Y.

(b) Are X and Y independent?

(a) P(X = – 1) = 0.04 + 0.02 + 0.06 + 0.12 = 0.24,

P(X = 0) = 0.50, P(X = 1) = 0.26

P(Y = 1) = 0.04 + 0.08 + 0.16 = 0.28,

P(Y = 2) = 0.16, P(Y = 3) = 0.30, P(Y = 5) = 0.26

(b) No, since P(X = – 1, Y = 1) ≠ P(X = – 1) P(Y = 1)

6. The probability of being able to log on to a certain computer from a remote terminal
at any given time is 0.7. Let X denote the number of independent attempts that must
be made to gain access to the computer. Find the PMF of X.

P(X = 1) = 0.7
P(X = 2) = P(1st = Fail, 2nd = Success) = P(1st = Fail) P(2nd = Success) = (0.3) (0.7) =
0.21
P(X = 3) = P(FFS) = (0.3)2 (0.7) = 0.063, and so on.

In general, for any positive integer k, P(X = k) = (0.3)(k – 1) (0.7).

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7. Every day, the number of network blackouts has a distribution (PMF) P(0)=0.6,
P(1)=0.2, P(2)=0.2, independently of other days. What is the probability that there
are more blackouts on Friday than on Thursday?

Let F = # blackouts on Friday, T = # blackouts on Thursday.

Using the theorem of total probability,

P(F > T) = P(F > T, T = 0) + P(F > T, T = 1) + P(F > T, T = 2)
= P(F > 0, T = 0) + P(F > 1, T = 1) + P(F > 2, T = 2)
= {P(F = 1, T = 0) + P(F = 2, T = 0)} + P(F = 2, T =1) + 0
= P(F = 1) P(T = 0) + P(F = 2) P(T = 0) + P(F = 2) P(T = 1) [independence]
= (0.2) (0.6) + (0.2) (0.6) + (0.2) (0.2)
= 0.28

Another way to think about it: Make a table of joint PMF for F and T, and observe that
the event {F > T} consists of the outcomes {(F=1, T=0), (F=2, T=0), (F=2, T=1)}.

8. An internet search engine looks for a certain keyword in a sequence of independent

web sites. It is believed that 20% of the sites contain this keyword.

(a) Let X be the number of websites visited until the first keyword is found. Find the
PMF of X.

P(X = x) = P(First x – 1 sites don’t have the keyword, x-th site has it)
= (0.80)(x – 1) (0.20), for x = 1, 2, 3, ….

(b) Out of the first 5 web sites, let Y be the number of sites that contain the keyword.
Find the PMF of Y.

P(Y = 0) = P(None of the sites has the keyword) = (0.80) 5 = 0.3277

 5
P(Y = 1) = P(1 site has the keyword, 4 sites don’t) =  (0.20)(0.80) 4 = 0.4096.
1
Remember to multiply by “5 choose 1” as there are “5 choose 1 = 5” outcomes where
1 site has the keyword and the other 4 sites don’t.

 5
P(Y = 2) = P(2 sites have, 3 sites don’t) =  (0.20) 2 (0.80) 3 = 0.2048
 2
 5
P(Y = 3) = P(3 sites have, 2 sites don’t) =  (0.20) 3 (0.80) 2 = 0.0512
 3
5
P(Y = 4) = P(4 sites have, 1 site doesn’t) =  (0.20) 4 (0.80) = 0.0064
 4

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P(Y = 5) = P(All sites have) = (0.20) 5 = 0.0003

(c) Compute the probability that at least 3 of the first 5 websites contain the keyword.

P(Y ≥ 3) = 0.0512 + 0.0064 + 0.0003 = 0.0579

(d) Compute the probability that the search engine had to visit at least 5 sites in order
to find the first occurrence of a keyword.

P(X ≥ 5) = 1 – P(X ≤ 4) = 1 – 0.5904 = 0.4096

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