4: Probability and Probability Distributions



Stat501 Homework #2

3.42 a Use MINITAB to compute

[pic].

Attach output!

b-c The scatterplot should look like the one shown below. Since the correlation coefficient is close to [pic], there is a fairly strong negative trend. This is quite an unexpected relationship!

[pic]

3.43 a Use Minitab to get

[pic].

b-c The scatterplot should look like the one shown on the next page. The correlation coefficient should be close to [pic]. There is a strong positive trend.

[pic]

3.45

[pic]

a The slope is [pic], the change in y for a one-unit change in x.

b As x increases, so does y, as indicated by the positive slope.

c When [pic]. This is the y-intercept = a.

d When [pic] When [pic]

4.6 a The experiment consists of selecting one of 25 students and recording the student’s gender as well as whether or not the student had gone to preschool.

b The experiment is accomplished in two stages, as shown in the tree diagram below.

Gender Preschool Simple Events Probability

Yes E1 : Male, Preschool 8/25

Male

No E2 : Male, No preschool 6/25

Yes E3 : Female, Preschool 9/25

Female

No E4 : Female, No preschool 2/25

c Since each of the 25 students are equally likely to be chosen, the probabilities will be proportional to the number of students in each of the four gender-preschool categories. These probabilities are shown in the last column of the tree diagram above.

d [pic]

4.9 The four possible outcomes of the experiment, or simple events, are represented as the cells of a [pic]table, and have probabilities as given in the table.

a P[adult judged to need glasses] = .44 + .14 = .58

b P[adult needs glasses but does not use them] = .14

c P[adult uses glasses] = .44 + .02 = .46

4.11 a Experiment: Select three people and record their gender (M or F).

b Extend the tree diagram in Figure 4.3 of the text to include one more coin toss (a total of n = 3). Then replace the H and T by M and F to obtain the 8 possible simple events shown below:

FFF FMM MFM MMF

MFF FMF FFM MMM

c Since there are N = 8 equally likely simple events, each is assigned probability, [pic].

d-e Sum the probabilities of the appropriate simple events:

[pic]

4.15 Similar to Exercise 4.9. The four possible outcomes of the experiment, or simple events, are represented as the cells of a [pic]table, and have probabilities (when divided by 300) as given in the table.

a [pic]

b [pic]

c [pic]

4.28 Use the extended mn Rule. The total number of options are [pic]

d Again, let A be the event of interest. There are [pic]ways (from part b) for the event A to occur, out of a total of [pic]possible configurations from part a, and the probability of interest is

[pic]

4.33 Notice that a sample of 10 nurses will be the same no matter in which order they were selected. Hence, order is unimportant and combinations are used. The number of samples of 10 selected from a total of 90 is

[pic]

4.34 a Since the order of selection for the two-person subcommittee is unimportant, use combinations to find the number of possible subcommittees as [pic]

b Since there is only [pic]subcommittee made up of Smith and Jones, the probability of this choice is [pic].

4.47 Refer to the solution to Exercise 4.1 where the six simple events in the experiment are given, with [pic].

a [pic] and [pic]

b [pic]

c [pic] and [pic]

d [pic]contains no simple events, and [pic]

e [pic]

f [pic]contains no simple events, and [pic]

g [pic]contains no simple events, and [pic]

h [pic] and [pic]

i [pic]and [pic]

4.59 Fix the birth date of the first person entering the room. Then define the following events:

A2: second person’s birthday differs from the first

A3: third person’s birthday differs from the first and second

A4: fourth person’s birthday differs from all preceding

[pic]

An: nth person’s birthday differs from all preceding

Then

[pic]

since at each step, one less birth date is available for selection. Since event B is the complement of event A,

[pic]

a For [pic], [pic]

b For [pic], [pic]

4.64 a Each of the four cell events is equally likely with [pic]. Hence, the probability of at least one dominate allele is

[pic]

b Similar to part a. The probability of at least one recessive allele is

[pic]

c Define the events: A: plant has red flowers

B: plant has one recessive allele

Then

[pic]

4.72 Define the following events: V: crime is violent

R: crime is reported

It is given that [pic]

a The overall reporting rate for crimes is

[pic]

b Use Bayes’ Rule: [pic]

and [pic]

c Notice that the proportion of non-violent crimes (.8) is much larger than the proportion of violent crimes (.2). Therefore, when a crime is reported, it is more likely to be a non-violent crime.

4.78 Define the following events, under the assumptions that an incorrect return has been filed.

G1: individual guilty of cheating

G2: individual not guilty (filed incorrectly due to lack of knowledge)

D: individual denies knowledge of error

It is given that [pic] Note that [pic]since if the individual has incorrectly filed due to lack of knowledge, he will, with probability 1 deny knowledge of the error. Using Bayes’ Rule,

[pic]

4.79 a Using the probability table,

[pic] [pic]

[pic] [pic]

b Using Bayes’ Rule, [pic]

c Using the definition of conditional probability,

[pic]

d [pic]

e [pic]

f The probability of a false negative is quite high, and would cause concern about the reliability of the screening method.

4.85 For the probability distribution given in this exercise,

[pic].

4.87 a-b On the first try, the probability of selecting the proper key is 1/4. If the key is not found on the first try, the probability changes on the second try. Let F denote a failure to find the key and S denote a success. The random variable is x, the number of keys tried before the correct key is found. The four associated simple events are shown below.

E1: S [pic] E3: FFS [pic]

E2: FS [pic] E4: FFFS [pic]

c-d Then

[pic]

The probability distribution and probability histogram follow.

|x |1 |2 |3 |4 |

|p(x) |1/4 |1/4 |1/4 |1/4 |

[pic]

4.90 Similar to Exercise 4.89. The random variable x can take on the values 0, 1, or 2. The associated probabilities can be found by summing probabilities of the simple events for the respective numerical events or by using the laws of probability:

[pic]

[pic]

[pic]

The probability distribution for x and the probability histogram follow.

|x |0 |1 |2 |

|p(x) |1/5 |3/5 |1/5 |

[pic]

4.91 Let x be the number of drillings until the first success (oil is struck). It is given that the probability of striking oil is [pic], so that the probability of no oil is [pic]

a [pic]

[pic]. This is the probability that oil is not found on the first drilling, but is found on the second drilling. Using the Multiplication Law,

[pic].

Finally, [pic].

b-c For the first success to occur on trial x, (x – 1) failures must occur before the first success. Thus,

[pic]

since there are (x – 1) N’s in the sequence. The probability histogram is shown below.

[pic]

4.92 Consider the event [pic]. This will occur only if either A or B wins three sets in a row; that is, if the event AAA or BBB occurs. Then

[pic]

Consider the event [pic]. This will occur if 3 A-wins are spread over 4 sets (with the last A-win in the fourth set) or if 3 B-wins are similarly spread over 4 sets. The associated simple events are

ABAA BABB AABA

BAAA ABBB BBAB

and the probability that [pic]is

[pic]

The event [pic]will occur if 4 A-wins are spread over 5 sets (with the last A-win in the fifth set) or similarly for B. The associated simple events are

ABBAA AABBA BBAAA BAABA ABABA BABAA

ABBAB AABBB BBAAB BAABB ABABB BABAB

and the probability that [pic]is

[pic]

Notice that [pic].

4.97 a Refer to the probability distribution for x given in this exercise.

[pic]

b [pic]

c [pic]

[pic] and [pic].

d Calculate [pic]or (1.078 to 3.718. Then, referring to the probability distribution of x, [pic].

4.102 a This experiment consists of two patients, each swallowing one of four tablets (two cold and two aspirin). There are four tablets to choose from, call them C1, C2, A1 and A2. The resulting simple events are then all possible ordered pairs which can be formed from the four choices.

(C1C2) (C2C1) (A1C1) (A2C1)

(C1A1) (C2A1) (A1C2) (A2C2)

(C1A2) (C2A2) (A1A2) (A2A1)

Notice that it is important to consider the order in which the tablets are chosen, since it makes a difference, for example, which patient (A or B) swallows the cold tablet.

b [pic]

c [pic]

d [pic]

4.107 The random variable x, defined as the number of householders insured against fire, can assume the values 0, 1, 2, 3 or 4. The probability that, on any of the four draws, an insured person is found is .6; hence, the probability of finding an uninsured person is .4. Note that each numerical event represents the intersection of the results of four independent draws.

1 [pic], since all four people must be uninsured.

2 [pic] (Note: the 4 appears in this expression because [pic]is the union of four mutually exclusive events. These represent the 4 ways to choose the single insured person from the fours.)

3 [pic], since the two insured people can be chosen in any of 6 ways.

4 [pic] and [pic].

Then

[pic]

4.108 In this exercise, x may take the values 0, 1, 2, or 3, and the probabilities associated with x are evaluated as in Exercise 4.107.

a [pic]

The reader may verify that the probabilities sum to one and that [pic]for [pic]. The requirements for a probability distribution have been satisfied.

b The alarm will function if [pic]. Hence,

[pic]

c [pic]

[pic]

4.110 The completed table is shown below, and each of the possible pairings are equally likely with probability 1/16.

|ss yy |Ss yY |ssYy |ss YY |

|sS yy |sS yY |sS Yy |sS YY |

|Ss yy |Ss yY |Ss Yy |Ss YY |

|SS yy |SS yY |SS Yy |SS YY |

a Smooth yellow peas result from all pairing having at least one S and at least one Y. Hence,[pic].

b Smooth green peas result when the pairing has at least one S and the pair yy. Hence,[pic].

c Wrinkled yellow peas result when the pairing has at least one Y and the pair ss. Hence, [pic]

d Wrinkled green peas result only when the pairing is ss yy. Hence, [pic].

e Define: A: offspring has smooth yellow peas

B: offspring has one s allele

C: offspring has one s allele and one y allele

Then [pic]. Using the definition of conditional probability,

[pic] and [pic]

4.111 Similar to Exercise 4.7. An investor can invest in three of five recommended stocks. Unknown to him, only 2 out of 5 will show substantial profit. Let P1 and P2 be the two profitable stocks. A typical simple event might be (P1P2N3), which represents the selection of two profitable and one nonprofitable stock. The ten simple events are listed below:

E1: (P1P2N1) E2: (P1P2N2) E3: (P1P2N3) E4: (P2N1N2)

E5: (P2N1N3) E6: (P2N2N3) E7: (N1N2N3) E8: (P1N1N2)

E9: (P1N1N3) E10: (P1N2N3)

Then [pic]since the simple events are equally likely, with [pic]. Similarly,

[pic]

4.122 Define A: union strike fund is adequate to support a strike

C: union-management team makes a contract settlement within 2 weeks

It is given that [pic]. Then

[pic].

Since [pic]and [pic], it appears that the settlement of the contract is independent of the ability of the union strike fund to support the strike.

4.124 Let y represent the value of the premium which the insurance company charges and let x be the insurance company’s gain. There are four possible values for x. If no accident occurs or if an accident results in no damage to the car, the insurance company gains y dollars. If an accident occurs and the car is damaged, the company will gain either [pic]dollars, [pic]dollars, or [pic]dollars, depending upon whether the damage to the car is total, 60% of market value, or 20% of market value, respectively. The following probabilities are known.

[pic]

Hence,

[pic]

Similarly,

[pic]and [pic]

The gain x and its associated probability distribution are shown below. Note that p(y) is found by subtraction.

x p(x)

[pic] .012

[pic] .018

[pic] .12

y .85

Letting the expected gain equal zero, the value of the premium is obtained.

[pic]

4.126 This exercise provides an example of a lot acceptance sampling plan. Seven items are drawn from a large lot of bearings and we wish to calculate the probability of accepting the lot; that is, the probability of observing no defectives in the sample of seven. In order to obtain P[acceptance], it is necessary to assume that the lot is large enough so that the probability of getting a defective is not noticeably affected by repeated draws. For example, consider a lot which contains 10,000 items, 5000 of which are defective. The probability of getting a defective on the first draw is 5000/10,000 or 1/2 . Assume that a defective has been drawn. Then the probability of getting a defective on the second draw is 4999/9999, which is not noticeably different from 1/2 . Define the following events:

D: draw a defective

G: draw a nondefective, where [pic]and [pic]

A: the lot is accepted

In each case, the desired probability is

[pic]

If all the items in the lot are nondefective, then [pic]and the probability of acceptance is [pic]. If 1/10 are defective, then [pic] If 1/2 are defective, then [pic]

4.133 a Define P: shopper prefers Pepsi and C: shopper prefers Coke. Then if there is actually no difference in the taste, P(P) = P(C) = 1/2 and

[pic]

b

[pic]

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