Homework 01

Xingzhou Luo

1.3 Of Great importance to residents of central Florida is the amount of radioactive material present

in the soil of reclaimed phosphate mining areas. Measurements of the amount of 238U in 25 soil

samples were as follow (measurements in picocuries per gram):

0.74 6.47 1.90 2.69 0.75

0.32 9.99 1.77 2.41 1.96

1.66 0.70 2.42 0.54 3.36

3.59 0.37 1.09 8.32 4.06

4.55 0.76 2.03 5.70 12.48

Construct a relative frequency histogram for these data.

1.5 Given here is the relative frequency histogram associated with grade point averages (GPAs) of a

sample of 30 students:

a Which of the GPA categories identified on the horizontal axis are associated with the

largest proportion of students?

b What proportion of students had GPAs in each of the categories that you identified?

c What proportion of the students had GPAs less than 2.65?

a 2.45 – 2.65, 2.65 – 2.85

b 7/30

c 3/30 + 3/30 + 3/30 + 7/30 = 16/30

1.7 The self-reported heights of 105 students in a biostatistics class were used to construct the

histogram given below.

a Describe the shape of the histogram.

b Does this histogram have an unusual feature?

c Can you think of an explanation for the two peaks in the histogram? Is there some

consideration other than height that results in the two separate peaks? What is it?

a The histogram has two peaks and has lower values on both ends and in the center.

b Yes. A graph of normal distribution only has one peak in the center and goes lower to

both sides. This graph has two peaks with lower value in the center.

c Since the average heights for male differs from the female’s, when combining these two

groups of data in one graph, peaks do not overlap with each other, resulting two peaks

and lower value in the center.

1.9 Resting breathing rates for college-age students are approximately normally distributed with

mean 12 and standard deviation 2.3 breaths per minute. What fraction of all college-age

students have breathing rates in the following intervals?

a 9.7 to 14.3 breaths per minute.

b 7.4 to 16.6 breaths per minute.

c 9.7 to 16.6 breaths per minute.

d Less than 5.1 or more than 18.9 breaths per minute.

Since,

a 9.7 – 14.3 = (12 - 2.3) – (12 + 2.3) → 68%

b 7.4 – 16.6 = (12 - 2*2.3) – (12 + 2*2.3) → 95%

c 9.7 – 16.6 = (12 - 2.3) – (12 + 2*2.3) → 68% + (95% - 68%)/2 = 81.5%

d Resting Breathing Rates → RBR

RBR < 5.1 & RBR > 18.9 =

RBR < (12 – 3*2.3) & RBR > (12 + 3*2.3) → 1 – 100% = 0

1.12 Using the result of Exercise 1.11 to calculate s for the n = 6 sample measurements 1, 4, 2, 1, 3,

and 3.

From Exercise 1.11,

And,

y ̅= (1 + 4 + 2 + 1 + 3 + 3) / 6 = 7/3 ≈ 2.333

s2 = 1 / (6 - 1) * [(1 – 7/3)2 + (4 – 7/3)2 + (2 – 7/3)2 + (1 – 7/3)2 + (3 - 7/3)2 + (3 – 7/3)2 ]

≈ 1.467

s = 1.211

1.22 Prove that the sum of the deviation of a set of measurements about their mean is equal to zero;

that is,

Let y = {y1 + y2 + y3 + … + yn-2 + yn-1 + yn}

Then y ̅= (y1 + y2 + y3 + … + yn-2 + yn-1 + yn) / n

The sum of y = ∑ 𝑦𝑖𝑛𝑖=1 = n * y̅

The sum of σ = ∑ (𝑦𝑖 − 𝑦 ) = 𝑛𝑖=1 ∑ 𝑦𝑖𝑛

𝑖=1 − ∑ �̅�𝑛𝑖=1 = (n * y̅) – (n * y)̅ = 0

Thus, ∑ (𝑦𝑖 − 𝑦 ) = 𝑛𝑖=1 0

Homework 02

Xingzhou Luo

2.6 From a survey of 60 students attending a university, it was found that 9 were living off campus,

36 were undergraduates, and 3 were undergraduates living off campus. Find the number of

these students who were

a undergraduates, were living off campus, or both.

b undergraduates living on campus.

c graduate students living on campus.

Undergraduates: 36

Graduate: 60 – 36 = 24

On Campus: 60 – 9 = 51

Off Campus: 9

Undergraduates Graduates Total Students

On Campus 33 18 51

Off Campus 3 6 9

Total Students 36 24 60

a number of undergraduates: 36

number of students living off campus: 9

number of undergraduates student living off campus: 3

b number of undergraduates living on campus: 33

c number of graduates living on campus: 18

2.8 Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote

the set of all possible pairs that can be observed. [These pairs can be listed, for example, by

letting (2, 3) denote that a 2 was observed on the first die and a 3 on the second.]

a Define the following subsets of S:

A: The number on the second die is even.

B: The sum of the two numbers is even.

C: At least one number in the pair is odd.

b List the points in 𝐴, 𝐶, 𝐴 ∩ 𝐵, 𝐴 ∩ 𝐵, 𝐴 ∪ 𝐵, and 𝐴 ∩ 𝐶.

a A = {(1, 2), (2, 2), (3, 2), (4, 2), (5 ,2), (6, 2), (1, 4), (2, 4), (3, 4),

(4, 4), (5 ,4), (6, 4), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}

B = {(1, 1), (3, 1), (5, 1), (1, 3), (3, 3), (5, 3), (1, 5), (3, 5), (5, 5),

(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}

C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 5),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 3), (4, 5),

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 3), (6, 5)}

b 𝐴 = {(1, 2), (2, 2), (3, 2), (4, 2), (5 ,2), (6, 2), (1, 4), (2, 4), (3, 4),

(4, 4), (5 ,4), (6, 4), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}

𝐶 = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)

𝐴 ∩ 𝐵 = {(2, 2), (4, 2), (6, 2), (2, 4), (4, 4), (6, 4), (2, 6), (4, 6), (6, 6)}

𝐴 ∩ 𝐵 = {(1, 2), (3, 2), (5, 2), (1, 4), (3, 4), (5, 4), (1, 6), (3, 6), (5, 6)}

𝐴 ∪ 𝐵 = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 3), (2, 3), (3, 3),

(4, 3), (5, 3), (6, 3), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5),

(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}

𝐴 ∩ 𝐶 = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 3), (2, 3), (3, 3),

(4, 3), (5, 3), (6, 3), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}

2.12 A vehicle arriving at an intersection can turn right, turn left, or continue straight ahead. The

experiment consists of observing the movement of a single vehicle through the intersection.

a List the sample space for this experiment.

b Assuming that all sample points are equally likely, find the probability that the vehicle

turns.

Let 𝑆 denote sample space, 𝐸𝐿 denote turning left, 𝐸𝑅 denote turning right and 𝐸𝑆 denote going

straight.

a 𝑆 = {𝐸𝐿 , 𝐸𝑅 , 𝐸𝑆 }

b 𝑃(𝑡𝑢𝑟𝑛) = 𝑃(𝐸𝐿) + 𝑃(𝐸𝑅) = 1

3+

1

3=

2

3

2.21 If 𝐴 and 𝐵 are events, use the result derived in Exercise 2.5(a) and the Axioms in Definition

2.6 to prove that

𝑃(𝐴) = 𝑃(𝐴 ∩ 𝐵) + 𝑃(𝐴 ∩ 𝐵)

From Exercise 2.5(a):

𝐴 = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐵)

And Definition 2.6:

Suppose 𝐴 ∩ 𝐵 and 𝐴 ∩ 𝐵 are not mutually exclusive.

Then there exists a subset 𝑋 ⊆ 𝐴 ∩ 𝐵, 𝑋 ⊆ 𝐴 ∩ 𝐵 and 𝑋 ≠ ∅.

Then 𝑋 = {𝑥|(𝑥 ∈ 𝐴) ∩ (𝑥 ∈ 𝐵) ∩ (𝑥 ∈ 𝐵)}.

Since (𝑥 ∈ 𝐵) ∩ (𝑥 ∈ 𝐵) and 𝑋 can only be ∅, which contradicts with 𝑋 ≠ ∅.

Thus 𝐴 ∩ 𝐵 and 𝐴 ∩ 𝐵 are mutually exclusive.

𝑃(𝐴) = 𝑃 ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐵)) = 𝑃(𝐴 ∩ 𝐵) + 𝑃(𝐴 ∩ 𝐵)

2.34 A retailer sells only two styles of stereo consoles, and experience shows that these are in equal

demand. Four customers in succession come into the store to order stereos. The retailer is

interested in their preferences.

a List the possibilities for preference arrangements among the four customers (that is, list

the sample space).

b Assign probabilities to the sample points.

c Let A denote the event that all four customers prefer the same style. Find P(A).

Let 𝑆 denote sample space, 𝐸1 denote ordering one style of stereo console and 𝐸2 denote

ordering the other style of stereo console.

a 𝑆 = { (𝐸1, 𝐸1, 𝐸1, 𝐸1), (𝐸1, 𝐸1, 𝐸1, 𝐸2), (𝐸1, 𝐸1, 𝐸2, 𝐸1, ), (𝐸1, 𝐸1, 𝐸2, 𝐸2),

(𝐸1, 𝐸2, 𝐸1, 𝐸1), (𝐸1, 𝐸2, 𝐸1, 𝐸2), (𝐸1, 𝐸2, 𝐸2, 𝐸1), (𝐸1, 𝐸2, 𝐸2, 𝐸2),

(𝐸2, 𝐸1, 𝐸1, 𝐸1), (𝐸2, 𝐸1, 𝐸1, 𝐸2), (𝐸2, 𝐸1, 𝐸2, 𝐸1), (𝐸2, 𝐸1, 𝐸2, 𝐸2),

(𝐸2, 𝐸2, 𝐸1, 𝐸1), (𝐸2, 𝐸2, 𝐸1, 𝐸2), (𝐸2, 𝐸2, 𝐸2, 𝐸1), (𝐸2, 𝐸2, 𝐸2, 𝐸2)}

b 𝑃 =1

16

c 𝑃(𝐴) = 𝑃((𝐸1, 𝐸1, 𝐸1, 𝐸1) + 𝑃(𝐸2, 𝐸2, 𝐸2, 𝐸2)1

16+

1

16=

1

8

2.42 A personnel director for a corporation has hired ten new engineers. If three (distinctly different)

positions are open at a Cleveland plant, in how many ways can she fill the positions?

𝑃310 =

10!

(10 − 3)!= 720

2.48 If we wish to expand (𝑥 + 𝑦)8, what is the coefficient of 𝑥5𝑦3? What is the coefficient of 𝑥3𝑦5?

(𝑥 + 𝑦)0 = 1

(𝑥 + 𝑦)1 = 𝑥 + 𝑦

(𝑥 + 𝑦)2 = 𝑥2 + 2𝑥𝑦 + 𝑦2

(𝑥 + 𝑦)3 = 𝑥3 + 3𝑥2𝑦 + 3𝑥𝑦2 + 𝑦3

(𝑥 + 𝑦)4 = 𝑥4 + 4𝑥3𝑦 + 6𝑥2𝑦2 + 4𝑥𝑦3 + 𝑦4

……

(𝑥 + 𝑦)𝑛 = ( 𝑛𝑛−0

)𝑥𝑛𝑦0 + ( 𝑛𝑛−1

)𝑥𝑛−1𝑦1 + ( 𝑛𝑛−2

)𝑥𝑛−2𝑦2 + ⋯ + ( 𝑛𝑛−𝑛

)𝑥𝑛−𝑛𝑦𝑛

(𝑥 + 𝑦)𝑛 = ∑ (𝑛𝑖)𝑥𝑛−𝑖𝑦𝑖𝑛

𝑖=0

The coefficient of 𝑥5𝑦3 = (85) = 56

The coefficient of 𝑥5𝑦3 = (83) = 56

2.52 An experimenter wishes to investigate the effect of three variables—pressure, temperature,

and the type of catalyst—on the yield in a refining process. If the experimenter intends to

use three settings each for temperature and pressure and two types of catalysts, how many

experimental runs will have to be conducted if he wishes to run all possible combinations of

pressure, temperature, and types of catalysts?

3 Pressures, 3 temperatures, 2 catalysts

3 × 3 × 2 = 18

2.73 Gregor Mendel was a monk who, in 1865, suggested a theory of inheritance based on the

science of genetics. He identified heterozygous individuals for flower color that had two alleles

(one r = recessive white color allele and one R = dominant red color allele). When these

individuals were mated, 3/4 of the offspring were observed to have red flowers, and 1/4 had

white flowers. The following table summarizes this mating; each parent gives one of its alleles

to form the gene of the offspring.

We assume that each parent is equally likely to give either of the two alleles and that, if either

one or two of the alleles in a pair is dominant (R), the offspring will have red flowers. What is

the probability that an offspring has

a at least one dominant allele?

b at least one recessive allele?

c one recessive allele, given that the offspring has red flowers?

a 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑅) = 𝑃(𝑅𝑟) + 𝑃(𝑅𝑅) + 𝑃(𝑟𝑅) =1

4+

1

4+

1

4=

3

4

b 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑟) = 𝑃(𝑅𝑟) + 𝑃(𝑟𝑅) + 𝑃(𝑟𝑟) =1

4+

1

4+

1

4=

3

4

c 𝑃(𝑜𝑛𝑒 𝑟|𝑟𝑒𝑑) =𝑃(𝑜𝑛𝑒 𝑟∩𝑟𝑒𝑑)

𝑃(𝑟𝑒𝑑)=

2

43

4

=2

3

2.84 if 𝐴1, 𝐴2 and 𝐴3 are three events and 𝑃(𝐴1 ∩ 𝐴2) = 𝑃(𝐴1 ∩ 𝐴3) ≠ 0 but 𝑃(𝐴2 ∩ 𝐴3) = 0, show

that

𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝐴𝑖) = 𝑃(𝐴1) + 𝑃(𝐴2) + 𝑃(𝐴3) − 2𝑃(𝐴1 ∩ 𝐴2).

𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝐴1) = 𝑃(𝐴1) + 𝑃(𝐴2) + 𝑃(𝐴3) − 𝑃(𝐴1 ∩ 𝐴2) − 𝑃(𝐴1 ∩ 𝐴3) − 𝑃(𝐴2 ∩ 𝐴3)

= 𝑃(𝐴1) + 𝑃(𝐴2) + 𝑃(𝐴3) − 𝑃(𝐴1 ∩ 𝐴2) − 𝑃(𝐴1 ∩ 𝐴2) − 0

= 𝑃(𝐴1) + 𝑃(𝐴2) + 𝑃(𝐴3) − 2𝑃(𝐴1 ∩ 𝐴2)

2.85 If 𝐴 and 𝐵 are independent events, show that 𝐴 and 𝐵 are also independent. Are 𝐴 and 𝐵

independent?

Since 𝐴 and 𝐵 are independent events.

Then 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) × 𝑃(𝐵).

Then 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) − 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) − 𝑃(𝐴) × 𝑃(𝐵)

= 𝑃(𝐴) × (1 − 𝑃(𝐵)) = 𝑃(𝐴) × 𝑃(𝐵).

Thus 𝐴 and 𝐵 are independent events.

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴 ∪ 𝐵) = 1 − 𝑃(𝐴 ∪ 𝐵) = 1 − (𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵))

= 1 − 𝑃(𝐴) − 𝑃(𝐵) + 𝑃(𝐴) × 𝑃(𝐵) = 𝑃(𝐴) − (𝑃(𝐵) × (1 − 𝑃(𝐴)))

= 𝑃(𝐴) − 𝑃(𝐵) × 𝑃(𝐴) = 𝑃(𝐴) × (1 − 𝑃(𝐵)) = 𝑃(𝐴) × 𝑃(𝐵)

Thus 𝐴 and 𝐵 are independent events.

2.93 Two events 𝐴 and 𝐵 are such that 𝑃(𝐴) = 0.2, 𝑃(𝐵) = 0.3, and 𝑃(𝐴 ∪ 𝐵) = 0.4. Find the

following:

a 𝑃(𝐴 ∩ 𝐵)

b 𝑃(𝐴 ∪ 𝐵)

c 𝑃(𝐴 ∩ 𝐵)

d 𝑃(𝐴|𝐵)

a 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∪ 𝐵) = 0.2 + 0.3 − 0.4 = 0.1

b 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴 ∩ 𝐵) = 1 − 𝑃(𝐴 ∩ 𝐵) = 1 − 0.1 = 0.9

c 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴 ∪ 𝐵) = 1 − 𝑃(𝐴 ∪ 𝐵) = 1 − 0.4 = 0.6

d 𝑃(𝐴|𝐵) =𝑃(𝐴∩𝐵)

𝑃(𝐵)=

𝑃(𝐵)−𝑃(𝐴∩𝐵)

𝑃(𝐵)=

0.3−0.1

0.3=

2

3

2.100 Show that Theorem 2.6, the additive law of probability, holds for conditional probabilities.

That is, if 𝐴, 𝐵 and 𝐶 are events such that 𝑃(𝐶) > 0, prove that 𝑃(𝐴 ∪ 𝐵|𝐶) = 𝑃(𝐴|𝐶) +

𝑃(𝐵|𝐶) − 𝑃(𝐴 ∩ 𝐵|𝐶). [Hint: Make use of the distributive law (𝐴 ∪ 𝐵) ∩ 𝐶 = (𝐴 ∪ 𝐶) ∩(𝐵 ∩ 𝐶).]

𝑃(𝐴 ∪ 𝐵|𝐶) =𝑃(𝐴 ∪ 𝐵)

𝑃(𝐶)=

𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)

𝑃(𝐶)= 𝑃(𝐴|𝐶) + 𝑃(𝐵|𝐶) − 𝑃(𝐴 ∩ 𝐵|𝐶)

2.104 If 𝐴 and 𝐵 are two events, prove that 𝑃(𝐴 ∩ 𝐵) ≥ 1 − 𝑃(𝐴) − 𝑃(𝐵). [Note: This is a simplified

version of the Bonferroni inequality.]

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∪ 𝐵) = (1 − 𝑃(𝐴)) + (1 − 𝑃(𝐵)) − 𝑃(𝐴 ∪ 𝐵)

= 2 − 𝑃(𝐴) − 𝑃(𝐵) − 𝑃(𝐴 ∪ 𝐵) = (1 − 𝑃(𝐴) − 𝑃(𝐵)) + (1 − 𝑃(𝐴 ∪ 𝐵))

Since 𝑃(𝐴 ∪ 𝐵) ≤ 1, then 1 − 𝑃(𝐴 ∪ 𝐵) ≥ 0.

Thus 𝑃(𝐴 ∩ 𝐵) ≥ 1 − 𝑃(𝐴) − 𝑃(𝐵).

2.108 If 𝐴, 𝐵 and 𝐶 are three events, use two applications of the result in Exercise 2.104 to prove that

𝑃(𝐴 ∩ 𝐵 ∩ 𝐶) ≥ 1 − 𝑃(𝐴) − 𝑃(𝐵) − 𝑃(𝐶).

Suppose 𝑋 = 𝐴 ∩ 𝐵.

According to result in Exercise 2.104:

𝑃(𝐴 ∩ 𝐵) ≥ 1 − 𝑃(𝐴) − 𝑃(𝐵).

Then 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶) = 𝑃(𝑋 ∩ 𝐶) ≥ 1 − 𝑃(𝑋) − 𝑃(𝐶).

Since 1 − 𝑃(𝑋) = 𝑃(𝑋).

Then 1 − 𝑃(𝑋) − 𝑃(𝐶) = 𝑃(𝑋) − 𝑃(𝐶) = 𝑃(𝐴 ∩ 𝐵) − 𝑃(𝐶).

Since 𝑃(𝐴 ∩ 𝐵) ≥ 1 − 𝑃(𝐴) − 𝑃(𝐵).

Then 𝑃(𝐴 ∩ 𝐵) − 𝑃(𝐶) ≥ 1 − 𝑃(𝐴) − 𝑃(𝐵) − 𝑃(𝐶).

Thus 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶) ≥ 1 − 𝑃(𝐴) − 𝑃(𝐵) − 𝑃(𝐶).

Homework 03

Xingzhou Luo

2.132 Use Theorem 2.8, the law of total probability, to prove the following:

a If 𝑃(𝐴|𝐵) = 𝑃(𝐴|𝐵), then 𝐴 and 𝐵 are independent.

b If 𝑃(𝐴|𝐶) > 𝑃(𝐵|𝐶) and 𝑃(𝐴|𝐶) > 𝑃(𝐵|𝐶), then 𝑃(𝐴) > 𝑃(𝐵).

Theorem 2.8:

a Since 𝐵 and 𝐵 are partitions of 𝑆.

Then 𝑃(𝐴) = 𝑃(𝐴|𝐵)𝑃(𝐵) + 𝑃(𝐴|𝐵)𝑃(𝐵).

= 𝑃(𝐴|𝐵)𝑃(𝐵) + 𝑃(𝐴|𝐵)(1 − 𝑃(𝐵))

= 𝑃(𝐴|𝐵)𝑃(𝐵) + 𝑃(𝐴|𝐵) − 𝑃(𝐴|𝐵)𝑃(𝐵) = 𝑃(𝐴|𝐵).

Since 𝑃(𝐴) = 𝑃(𝐴|𝐵).

Thus 𝐴 and 𝐵 are independent.

b Since 𝐶 and 𝐶 are partitions of 𝑆.

Then 𝑃(𝐴) = 𝑃(𝐴|𝐶)𝑃(𝐶) + 𝑃(𝐴|𝐶)𝑃(𝐶).

And 𝑃(𝐵) = 𝑃(𝐵|𝐶)𝑃(𝐶) + 𝑃(𝐵|𝐶)𝑃(𝐶).

Since 𝑃(𝐴|𝐶) > 𝑃(𝐵|𝐶) and 𝑃(𝐴|𝐶) > 𝑃(𝐵|𝐶), 𝑃(𝐶) ≥ 0 and 𝑃(𝐶) ≥ 0.

Thus 𝑃(𝐴) > 𝑃(𝐵).

2.132 A plane is missing and is presumed to have equal probability of going down in any of three

regions. If a plane is actually down in region 𝑖, let 1 − 𝛼𝑖 denote the probability that the plane

will be found upon a search of the 𝑖th region, 𝑖 = 1, 2, 3. What is the conditional probability that

the plane is in

a region 1, given that the search of region 1 was unsuccessful?

b region 2, given that the search of region 1 was unsuccessful?

c region 3, given that the search of region 1 was unsuccessful?

Let 𝐹𝑖 denote the plane is searched and found in region 𝑖, and 𝑅𝑖 denote the plane is in region 𝑖.

𝑃(𝐹1|𝑅1) = 1 − 𝑎1 𝑃(𝑅1) =1

3

𝑃(𝐹2|𝑅2) = 1 − 𝑎2 𝑃(𝑅2) =1

3

𝑃(𝐹3|𝑅3) = 1 − 𝑎3 𝑃(𝑅3) =1

3

a 𝑃(𝑅1|𝐹1) =𝑃(𝑅1∩𝐹1)

𝑃(𝐹1)=

𝑃(𝐹1|𝑅1)𝑃(𝑅1)

∑ 𝑃(𝐹1|𝑅𝑖)𝑃(𝑅𝑖)3𝑖=1

=1

3𝑎1

1

3𝑎1+

1

3(𝑃(𝐹1|𝑅2)+𝑃(𝐹1|𝑅3))

Since 𝑃(𝐹1|𝑅2) is the probability of search and find the plane in region 1 given the plane

is in region 2, 𝑃(𝐹1|𝑅2) = 0, thus, 𝑃(𝐹1|𝑅2) = 1 − 𝑃(𝐹1|𝑅2) = 1.

Same applies to 𝑃(𝐹1|𝑅3), thus 𝑃(𝐹1|𝑅3) = 1.

𝑃(𝑅1|𝐹1) =

13𝑎1

13𝑎1 +

13(𝑃(𝐹1|𝑅2) + 𝑃(𝐹1|𝑅3))

=

13𝑎1

13𝑎1 +

23

=𝑎1

𝑎1 + 2

b 𝑃(𝑅2|𝐹1) =𝑃(𝑅2∩𝐹1)

𝑃(𝐹1)=

𝑃(𝐹1|𝑅2)𝑃(𝑅2)

∑ 𝑃(𝐹1|𝑅𝑖)𝑃(𝑅𝑖)3𝑖=1

=1

3𝑎2

1

3𝑎2+

1

3(𝑃(𝐹1|𝑅1)+𝑃(𝐹1|𝑅3))

For the same reason from above, 𝑃(𝐹1|𝑅1) + 𝑃(𝐹1|𝑅3) = 2.

𝑃(𝑅2|𝐹1) =𝑎2

𝑎2+2

c 𝑃(𝑅3|𝐹1) =𝑃(𝑅3∩𝐹1)

𝑃(𝐹1)=

𝑃(𝐹1|𝑅3)𝑃(𝑅3)

∑ 𝑃(𝐹1|𝑅𝑖)𝑃(𝑅𝑖)3𝑖=1

=1

3𝑎3

1

3𝑎3+

1

3(𝑃(𝐹1|𝑅1)+𝑃(𝐹1|𝑅2))

For the same reason from above, 𝑃(𝐹1|𝑅1) + 𝑃(𝐹1|𝑅2) = 2.

𝑃(𝑅3|𝐹1) =𝑎3

𝑎3+2

2.135 Of the travelers arriving at a small airport, 60% fly on major airlines, 30% fly on privately

owned planes, and the remainder fly on commercially owned planes not belonging to a major

airline. Of those traveling on major airlines, 50% are traveling for business reasons, whereas

60% of those arriving on private planes and 90% of those arriving on other commercially owned

planes are traveling for business reasons. Suppose that we randomly select one person arriving

at this airport. What is the probability that the person

a is traveling on business?

b is traveling for business on a privately owned plane?

c arrived on a privately owned plane, given that the person is traveling for business

reasons?

d is traveling on business, given that the person is flying on a commercially owned plane?

Major Airlines: 𝑃(𝑀) =3

5 𝑃(𝐵|𝑀) =

1

2

Private Planes: 𝑃(𝑃) =3

10 𝑃(𝐵|𝑃) =

3

5

Commercial Planes: 𝑃(𝐶) =1

10 𝑃(𝐵|𝐶) =

9

10

a 𝑃(𝐵) = 𝑃(𝐵|𝑀)𝑃(𝑀) + 𝑃(𝐵|𝑃)𝑃(𝑃) + 𝑃(𝐵|𝐶)𝑃(𝐶)

=3

5×

1

2+

3

10×

3

5+

1

10×

9

10= 0.3 + 0.18 + 0.09 = 0.57

b 𝑃(𝐵 ∩ 𝑃) = 𝑃(𝐵|𝑃)𝑃(𝑃) = 0.3 × 0.6 = 0.18

c 𝑃(𝑃|𝐵) =𝑃(𝑃∩𝐵)

𝑃(𝐵)=

0.18

0.57≈ 0.316

d 𝑃(𝐵|𝐶) = 0.9

3.2 You and a friend play a game where you each toss a balanced coin. If the upper faces on the

coins are both tails, you win $1; if the faces are both heads, you win $2; if the coins do not

match (one shows a head, the other a tail), you lose $1 (win (−$1)). Give the probability

distribution for your winnings, 𝑌, on a single play of this game.

𝑃(𝑌 = 1) =1

4

𝑃(𝑌 = 2) =1

4

𝑃(𝑌 = −1) =1

2

3.4 Consider a system of water flowing through valves from 𝐴 to 𝐵. (See the accompanying

diagram.) Valves 1, 2, and 3 operate independently, and each correctly opens on signal with

probability 0.8. Find the probability distribution for 𝑌, the number of open paths from 𝐴 to 𝐵

after the signal is given. (Note that 𝑌 can take on the values 0, 1, and 2.)

𝑃(𝑌 = 0):

𝑝(𝑛𝑜𝑛𝑒) = 𝑝(𝑃1 ∩ 𝑃2 ∩ 𝑃3) = 0.2 × 0.2 × 0.2 = 0.008

𝑝(𝑜𝑛𝑙𝑦 2) = 𝑝(𝑃1 ∩ 𝑃2 ∩ 𝑃3) = 0.2 × 0.8 × 0.2 = 0.032

𝑝(𝑜𝑛𝑙𝑦 3) = 𝑝(𝑃1 ∩ 𝑃2 ∩ 𝑃3) = 0.2 × 0.2 × 0.8 = 0.032

𝑃(𝑌 = 1):

𝑝(𝑜𝑛𝑙𝑦 1) = 𝑝(𝑃1 ∩ 𝑃2 ∩ 𝑃3) = 0.8 × 0.2 × 0.2 = 0.032

𝑝(𝑜𝑛𝑙𝑦 1 𝑎𝑛𝑑 2) = 𝑝(𝑃1 ∩ 𝑃2 ∩ 𝑃3) = 0.8 × 0.8 × 0.2 = 0.128

𝑝(𝑜𝑛𝑙𝑦 1 𝑎𝑛𝑑 3) = 𝑝(𝑃1 ∩ 𝑃2 ∩ 𝑃3) = 0.8 × 0.2 × 0.8 = 0.128

𝑝(𝑜𝑛𝑙𝑦 2 𝑎𝑛𝑑 3) = 𝑝(𝑃1 ∩ 𝑃2 ∩ 𝑃3) = 0.2 × 0.8 × 0.8 = 0.128

𝑃(𝑌 = 1):

𝑝(𝑎𝑙𝑙) = 𝑝(𝑃1 ∩ 𝑃2 ∩ 𝑃3) = 0.8 × 0.8 × 0.8 = 0.512

𝑃(𝑌 = 0) = 𝑝(𝑛𝑜𝑛𝑒) + 𝑝(𝑜𝑛𝑙𝑦 2) + 𝑝(𝑜𝑛𝑙𝑦 3) = 0.072

𝑃(𝑌 = 1) = 𝑝(𝑜𝑛𝑙𝑦 1) + 𝑝(𝑜𝑛𝑙𝑦 2 𝑎𝑛𝑑 3) + 𝑝(𝑜𝑛𝑙𝑦 1 𝑎𝑛𝑑 3) + 𝑝(𝑜𝑛𝑙𝑦 2 𝑎𝑛𝑑 3) = 0.416

𝑃(𝑌 = 2) = 𝑝(𝑎𝑙𝑙) = 0.512

3.9(a, c) In order to verify the accuracy of their financial accounts, companies use auditors on a regular

basis to verify accounting entries. The company’s employees make erroneous entries 5% of

the time. Suppose that an auditor randomly checks three entries.

a Find the probability distribution for 𝑌, the number of errors detected by the auditor.

b Construct a probability histogram for 𝑝(𝑦).

c Find the probability that the auditor will detect more than one error.

a 𝑃(𝑌 = 0) = (30) × 0.95 × 0.95 × 0.95 = 0.857375

𝑃(𝑌 = 1) = (31) × 0.05 × 0.95 × 0.95 = 0.135375

𝑃(𝑌 = 2) = (32) × 0.05 × 0.05 × 0.95 = 0.007125

𝑃(𝑌 = 3) = (33) × 0.05 × 0.05 × 0.05 = 0.000125

c 𝑃(𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑡 1) = 𝑃(𝑌 = 2) + 𝑃(𝑌 = 3) = 0.00725

3.12 Let 𝑌 be a random variable with 𝑝(𝑦) given in the accompanying table. Find 𝐸(𝑌), 𝐸(1/𝑌),

𝐸(𝑌2 − 1), and 𝑉(𝑌).

𝐸(𝑌) = ∑ 𝑦𝑝(𝑦)

𝑦

= 1 × 0.4 + 2 × 0.3 + 3 × 0.2 + 4 × 0.1 = 2

𝐸 (1

𝑌) = ∑

𝑝(𝑦)

𝑦𝑦 =0.4

1+

0.3

2+

0.2

3+

0.1

4=

77

120

𝐸(𝑌2 − 1) = ∑ (𝑦2 − 1)𝑝(𝑦) = 0 × 0.4 + 3 × 0.3 + 8 × 0.2 + 15 × 0.1 = 4𝑦

𝑉(𝑌) = 𝐸[(𝑌 − 𝜇)2] = 𝐸(𝑌2) − 𝜇2 = ∑ 𝑦2𝑝(𝑦) −𝑦 𝜇2

= 1 × 0.4 + 4 × 0.3 + 9 × 0.2 + 16 × 0.1 − 4 = 1

3.24 Approximately 10% of the glass bottles coming off a production line have serious flaws in the

glass. If two bottles are randomly selected, find the mean and variance of the number of bottles

that have serious flaws.

Let 𝑌 denote the event, then:

𝑦 0 1 2

𝑝(𝑦) (

2

0) ×

9

10×

9

10= 0.81 (

2

1) ×

1

10×

9

10= 0.18 (

2

0) ×

1

10×

1

10= 0.01

𝜇 = 𝐸(𝑌) = ∑ 𝑦𝑝(𝑦)

𝑦

= 0 × 0.81 + 1 × 0.18 + 2 × 0.01 = 0.2

𝜎2 = 𝑉(𝑌) = 𝐸[(𝑌 − 𝜇)2] = 𝐸(𝑌2) − 𝜇2 = ∑ 𝑦2𝑝(𝑦) −𝑦 𝜇2

= 0 × 0.81 + 1 × 0.18 + 4 × 0.01 − 0.04 = 0.18

3.34 The manager of a stockroom in a factory has constructed the following probability distribution

for the daily demand (number of times used) for a particular tool.

It costs the factory $10 each time the tool is used. Find the mean and variance of the daily cost

for use of the tool.

𝑐 0 10 20

𝑝(𝑐) 0.1 0.5 0.4

𝜇 = 𝐸(𝐶) = ∑ 𝑐𝑝(𝑐)

𝑦

= (0 × 0.1 + 10 × 0.5 + 20 × 0.4) = 13

𝜎2 = 𝑉(𝐶) = ∑ 𝑐2𝑝(𝑐) −𝑦 𝜇2

= (0 × 0.1 + 100 × 0.5 + 400 × 0.4) − 169 = 41

Homework 4

Xingzhou Luo

3.38 The manufacturer of a low-calorie dairy wishes to compare the taste appeal of a new formula

(formula 𝐵) with that of the standard formula (formula 𝐴). Each of four judges is given three

glasses in random order, two containing formula 𝐴 and the other containing formula 𝐵. Each

judge is asked to state which glass he or she most enjoyed. Suppose that the two formulas are

equally attractive. Ley 𝑌 be the number of judges stating a preference for the new formula.

a Find the probability function for 𝑌.

b What is the probability that at least three of the four judges state a preference for the new

formula?

c Fund the expected value of 𝑌.

d Find the variance of 𝑌.

a 𝑝(𝑦) = (4𝑦

) (1

3)

𝑦(

2

3)

4−𝑦, 𝑦 = 0, 1, 2, 3, 4.

b 𝑝(𝑦 > 2) = 𝑝(3) + 𝑝(4) = (43) (

1

3)

3(

2

3) + (4

4) (

1

3)

4=

1

9

c 𝐸(𝑌) = 4 ×1

3=

4

3

d 𝑉(𝑌) = 4 ×1

3×

2

3=

8

9

3.44 A new surgical procedure is successful with a probability of 𝑝. Assume that the operation is

performed five times and the results are independent of one another. What is the probability that

a all five operations are successful if 𝑝 = 0.8?

b exactly four are successful if 𝑝 = 0.6?

c less than two are successful if 𝑝 = 0.3?

a 𝑝(𝑌 = 5) = (55)𝑝5𝑞0 = 0.85 = 0.32768

b 𝑝(𝑌 = 4) = (54)𝑝4𝑞1 = 5 × 0.64 × 0.4 = 0.2592

c 𝑝(𝑌 < 2) = (51)𝑝1𝑞4 + (5

0)𝑝5𝑞0 = 5 × 0.3 × 0.74 + 0.75 = 0.52822

3.56 An oil exploration firm is formed with enough capital to finance ten explorations. The probability

of a particular exploration being successful is 0.1. Assume the explorations are independent. Find

the mean and variance of the number of successful explorations.

𝑝(𝑦) = (10𝑦

) × 0.1𝑦 × 0.910−𝑦

𝜇 = 𝐸(𝑌) = 𝑛𝑝 = 10 × 0.1 = 1

𝜎2 = 𝑉(𝑌) = 𝑛𝑝𝑞 = 10 × 0.1 × 0.9 = 0.9

3.70 An oil prospector will drill a succession of holes in a given area to find a productive well. The

probability that he is successful on a given trial is 0.2.

a What is the probability that the third hole drilled is the first to yield a productive well?

b If the prospector can afford to drill at most ten wells, what is the probability that he will

fail to find a productive well?

a 𝑝(1𝑠𝑡 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑓𝑢𝑙 𝑜𝑛 3𝑟𝑑 𝑑𝑟𝑖𝑙𝑙) = 0.82 × 0.2 = 0.128

b 𝑝(𝑛𝑜 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑓𝑢𝑙 𝑑𝑟𝑖𝑙𝑙) = 0.810 = 0.1073741824

3.80 Two people took turns tossing a fair die until one of them tossed a 6. Person A tossed first, B

second, A third, and so on. Given that person B threw the first 6, what is the probability that B

obtained the first 6 on her second toss (that is, on the fourth toss overall)?

𝑝(1𝑠𝑡 6 𝑜𝑛 4𝑡ℎ 𝑡𝑜𝑠𝑠) = (5

6)

3(

1

6) =

125

1296

𝑝(1𝑠𝑡 6 𝑜𝑛 𝑒𝑣𝑒𝑛 𝑡𝑜𝑠𝑠 ) = 1 − 𝑝(1𝑠𝑡 6 𝑜𝑛 𝑜𝑑𝑑 𝑡𝑜𝑠𝑠 )

= 1 −𝑝

1−𝑞2 = 1 −1

6

1−(5

6)

2 =5

11

𝑝(1𝑠𝑡 6 𝑜𝑛 4𝑡ℎ 𝑡𝑜𝑠𝑠 |1𝑠𝑡 6 𝑜𝑛 𝑒𝑣𝑒𝑛 𝑡𝑜𝑠𝑠) =𝑝(1𝑠𝑡 6 𝑜𝑛 4𝑡ℎ 𝑡𝑜𝑠𝑠)∩𝑝(1𝑠𝑡 6 𝑜𝑛 𝑒𝑣𝑒𝑛 𝑡𝑜𝑠𝑠)

𝑝(1𝑠𝑡 6 𝑜𝑛 𝑒𝑣𝑒𝑛 𝑡𝑜𝑠𝑠)

=𝑝(1𝑠𝑡 6 𝑜𝑛 4𝑡ℎ 𝑡𝑜𝑠𝑠)

𝑝(1𝑠𝑡 6 𝑜𝑛 𝑒𝑣𝑒𝑛 𝑡𝑜𝑠𝑠)

= 0.212

3.112 Used photocopy machines are returned to the supplier, cleaned, and then sent back on lease

agreements. Major repairs are not made, however, and as a result, some customers receive

malfunctioning machines. Among eight used photocopiers available today, three are

malfunctioning. A customer wants to lease four machines immediately. To meet the customer’s

deadline, four of the eight machines are randomly selected and, without further checking, shipped

to the customer. What is the probability that the customer receives

a no malfunctioning machines?

b at least one malfunctioning machine?

a 𝑝(𝑛𝑜 𝑚𝑎𝑙𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛) = (54) ÷ (8

4) =

1

14

b 𝑝(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑚𝑎𝑙𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛) = 1 − 𝑝(𝑛𝑜 𝑚𝑎𝑙𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛) = 1 −1

14=

13

14

Bonus Prove that let 𝑌 be a binomial random variable based on 𝑛 trials and success probability 𝑝. Then

𝜎2 = 𝑉(𝑌) = 𝑛𝑝𝑞.

𝑉(𝑌) = 𝐸[(𝑌 − 𝜇)2] = 𝐸(𝑌2) − 𝜇2 = 𝐸[𝑌(𝑌 − 1)] + 𝐸(𝑌) − [𝐸(𝑌)]2

= [∑ 𝑦(𝑦 − 1)𝑛!

𝑦!(𝑛−𝑦)!𝑛𝑦=0 𝑝𝑦𝑞𝑛−𝑦] + 𝑛𝑝 − 𝑛2𝑝2

= [∑ 𝑛(𝑛 − 1)𝑝2 (𝑛−2)!

(𝑦−2)!(𝑛−𝑦)!𝑛𝑦=0 𝑝𝑦−2𝑞𝑛−𝑦] + 𝑛𝑝 − 𝑛2𝑝2

= [𝑛(𝑛 − 1)𝑝2 ∑ (𝑛−2𝑦−2

) 𝑝𝑦−2𝑝𝑛−𝑦𝑛𝑦=0 ] + 𝑛𝑝 − 𝑛2𝑝2

Since 𝑝(𝑦 − 2) = (𝑛−2𝑦−2

) 𝑝𝑦−2𝑝𝑛−𝑦 is a binomial probability function based on 𝑛 − 2 trials.

Then ∑ 𝑝(𝑦 − 2) = 1𝑛𝑦=0 .

Thus 𝑉(𝑌) = 𝑛(𝑛 − 1)𝑝2 + 𝑛𝑝 + 𝑛2𝑝2

= 𝑛2𝑝2 − 𝑛𝑝2 + 𝑛𝑝 − 𝑛2𝑝2

= 𝑛𝑝(1 − 𝑝) = 𝑛𝑝𝑞