Stat 330
Exam I Practice Questions: Solutions
1. Short answers
(a) Give an example demonstrating why the number of combinations is always less than the number
of permutations.
n!
n!
P (n, k) = (n−k)!
and C(n, k) = nk = k!(n−k)!
e.g. P (4, 2) = 12 and C(4, 2) = 6
(b) List three axioms of probability.
These are the Kolmogorov Axioms:
i. for all events A, P (A) ≥ 0
ii. for all events A, B with A ∩ B = ∅: P (A ∪ B) = P (A) + P (B).
iii. P (Ω) = 1
(c) Give two examples which require a continuous sample space.
e.g. lifetime of an electrical device, height of a person, ...
(d) How does the geometric distribution differ from the Binomial?
a random variable with a geometric distribution counts the number of attempts until the first success, a random variable with a binomial distribution counts the number of successes in n repetitions
of the experiment.
2. Counting Examples
(a) A computer shops advertises computers with 5 choices of monitors, three sizes of hard disks and
2 sizes of main memory. At least one computer for each configuration is in the shop. How many
computers have to be in the shop at least? Explain.
5 · 3 · 2 = 30 different computers at least in the shop
(b) In an exam there are 10 questions, each candidate has to choose 3 questions.
How many possibilities to choose three questions are there
i. in total? Explain.
10
10 · 9 · 8
=
= 120.
3
3·2·1
ii. if at least one question has to be chosen from the first 4 questions? Explain.
4 6
1 2
| {z }
+
4 6
2 1
| {z }
+
4 6
3 0
| {z }
one from first 4.
two from first 4,
three from first 4,
two from second 6
one from second 6
none from second 6
alternatively,
10
−
3
| {z }
total, see i)
4 6
0 3
| {z }
none from first 4, three from second 6
= 4 · 15 + 6 · 6 + 4 = 100.
= 120 − 20 = 100.
(c) Someone owns 7 CDs with classical music, 12 CDs with pop music and 5 CDs with Jazz. How
many possibilities are there to put all CDs on a shelf, if all CDs of the same category are next to
each other? Explain.
· |{z}
7! ·
3!
|{z}
12!
|{z}
re-order 3 topics classical pop-music
· |{z}
5! ≈ 1.7382 · 1015 .
Jazz
(d) A group of 11 persons consists of 7 men and 4 women. 5 persons are selected for a committee.
How many possibilities are there
i. in total? Explain.
11
= 462.
5
ii. if exactly one women must be in the committee? Explain.
4
1
|{z}
7
·
= 4 · 35 = 140.
4
|{z}
one woman four men
iii. if at least one women must be in the committee? Explain.
−
462
|{z}
total number
7
5
|{z}
= 462 − 21 = 441.
no woman
(e) An administrator demands from his users that passwords must be chosen according to the following
rules:
• a password must have exactly 8 characters
• at least one character must be a digit ’0’-’9’ or one of the special characters ’#’,’-’,’ ’,’&’,’%’.
How many possible passwords are there? Explain.
We’ll assume, that we can take from the ’usual’ characters ’a’-’z’ and ’A’-’Z’ (that’s 52 characters,
or 26 , if we don’t distinguish between upper and lower case). In total, we have therefore 52+10+5
= 67 different characters.
678
|{z}
total number
−
528
|{z}
≈ 3.526079 · 1014 .
no special character
3. Some jobs submitted for processing on a particular CPU have fatal programming errors, while others
do not. Suppose that the long run fraction of jobs with fatal programming errors is p = 0.05.
a) Find (under appropriate distribution assumptions) the probability that among the next 10 jobs
submitted there are less than 3 with fatal errors.
The next 10 jobs can be considered 10 iid Bernoulli trials, with “a fatal error” considered a
“Success” and P (S) = .05 = p
Let X = # of jobs with a fatal error in the next 10. Then X ∼ Binomial(n = 10, p = .05)
P (X < 3)
=
P (X = 0) + P (X = 1) + P (X = 2)
10
10
10
=
(.05)0 (.95)1 0 +
(.05)1 (.95)9 +
(.05)2 (.95)8
0
1
2
= .59874 + 10 × .05 × .63025 + 45 × .0025 × .66342
= .9885
b) One begins monitoring the processing of jobs and lets
X = the total number of jobs processed until the first with a fatal error.
Find (again under appropriate distribution assumptions) P (X ≤ 30)
Here, obviously, X ∼ Geometric(p = .05)
P (X ≥ 4)
=
P (X > 3)
= P (there are 3 jobs without a fatal error
(.95)3 = .857375
=
4. In an experiment you toss two dice. A random variable X is defined as the minimum score of the two
dice.
a) Find the image of X.
im(X) = {1, 2, 3, 4, 5, 6}.
b) What is the probability that the minimum of the two dice is 3?
By looking at a sample space Ω with equally likely elementayr events we get the probability by just
counting


(1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6) 






(2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6) 






(3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6)
Ω=
(4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6) 






(5,
1),(5, 2),(5, 3),(5, 4),(5, 5),(5, 6) 






(6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6, 6)
2nd
1
2
3
4
5
6
1
1
1
1
1
1
1
2
1
2
2
2
2
2
P (X = 3) =
first
3
1
2
3
3
3
3
die
4
1
2
3
4
4
4
5
1
2
3
4
5
5
6
1
2
3
4
5
6
7
36
c) What is the probability that the minimum of the two dice is less than 3?
P (X < 3) = P (X = 1) + P (X = 2) =
11
9
10
+
=
.
36 36
18
d) Give the probability mass function of X. Explain
using the above sample space,
x
pX (x)
1
2
3
4
5
6
11
36
9
36
7
36
5
36
3
36
1
36
e) Draw probability mass and probability distribution function of X into the two boxes below. Do
not forget to mark and label the axes appropriately.
5. Given is the following table:
x
pX
−3
0.1
−1
0.15
0
0.3
1
s
(a) Find s, so that pX is a probability mass function.
s = 1 − (.1 + .15 + .3) = .45
(b) Assume, the random variable X has probability mass function pX . What is E[X]?
E[X] = −3(.1) − 1(.15) + 0(.3) + 1(.45) = 0
(c) Determine V ar[X]. E[X 2 ] = 9(.1) + 1(.15) + 0(.3) + 1(.45) = 1.5
V ar[X] = E[X 2 ] − (E[X])2 = 1.5
(d) Assume, Y is another random variable with E[Y ] = 1.25 and V ar[Y ] = 2.5. Evaluate the following
expressions:
i. E[X + 4Y ] E[X] + E[Y ] = 0 + 4(1.25) = 5.0
ii. V ar[3Y ] V ar[3Y ] = 32 V ar[Y ] = 9(2.5) = 22.5
iii. E[2Y 2 ] 2E[Y 2 ] = 2[V ar[Y ] + (E[Y ])2 ] = 2(2.5 + 1.252 ) = 8.125
6. In three boxes there are capacitors as shown in the following table:
Capacitance Number in box
(in µF )
1
2
3
1.0
10 90
25
0.1
50 30
80
70 90
120
0.01
An experiment consists of first randomly selecting a box (assume that each box has the same probability
of selection) and the randomly selecting a capacitor from the chosen box.
a) Draw a tree diagram and determine the probability of selecting a 0.1 µF capacitor.
box 1
1/13
1 F
5/13
0.1 F
7/13
0.01 F
1/3
3/7
1/3
box 2
1/3
1/7
3/7
1/9
box 3
1 F
0.1 F
0.01 F
1 F
80/225 0.1 F
120/225
0.01 F
P ( select 0.1 µF ) = 1/3 · 5/13 + 1/3 · 1/7 + 1/3 · 80/225 = 0.29.
b) If a 0.1 µF capacitor is chosen, what is the probability that it came from box 1?
P ( box 1 | 0.1µF ) =
P (0.1µF | box 1) · P ( box 1 )
5/13 · 1/3
=
= 0.44.
P (0.1µF )
0.29
Now assume, that the contents of all three boxes are poured into one big box.
c) What is now the probability of selecting a 0.1 µF capacitor?
P ( select 0.1 µF ) =
160
50 + 30 + 80
=
= 0.28.
130 + 210 + 225
565
d) If three capacitors are selected at random, what is the probability that at least one of them is a
0.1 µF capacitor?
In total there are 565 capacitors, 160 of them have 0.1 µF and 405 others.
P ( at least one 0.1 µF ) = 1 − P ( none ) = 1 −
405 · 404 · 403
= 1 − 0.3675 = 0.6325.
565 · 564 · 563
7. A manufacturing process produces integrated circuit chips. The fraction of bad chips produced is 20%.
Thoroughly testing a chip to determine whether it is good or bad is expensive, so a cheap test is tried.
All good chips will pass the cheap test, but so will 10% of the bad chips.
(a) Given a chip passes the cheap test, what is the probability that it is a good chip?
P (good|pass)
=
=
P (pass|good)P (good)
P (pass|good)P (good) + P (pass|bad)P (bad)
(1)(.8)
= .976
(1)(.8) + (.1)(.2)
(b) If a company using this manufacturing process sells all chips which pass the cheap test, what
percentage of chips sold will be bad?
P (bad|sold) = P (bad|pass) = 1 − P (good|pass) = .024