Solutions to Assignment 4

The problems you should study for the final exam are marked with a (***).

***1. Classify each of the following statements as an example of theoretical probability, relative frequency interpretation of probability, or personal (subjective) probability.

a. According to company records the probability that a washing machine will need repairs during a six year period is 0.10.

Because of the phrase "according to company records," the sentence suggests that the probability figure was based on the relative frequency of repairs.

b. The probability of choosing six number from 1 to 40 that match the six numbers drawn by a state lottery is 1/3,838,380 » 0.00000026.

All 3,838,380 possibilities are viewed as equally likely on theoretical grounds.

c. If Rex Grossman plays well, the Bears have a 75% chance of making it to the Super Bowl.

This statement is an example of subjective probability, a person's assessment of likelihood, not based on any repeated trials.  In fact, the situation in question is not repeatable.  In a different year, it would be a different Bears team against different opponent teams, etc.

d. We think there is a 50-50 chance of a recession next year.

This statement is another example of subjective probability.

e. The probability of a person from the United States being left handed is 0.11.

This statement suggests that it was based on a census or survey of some kind; hence, most likely it is an example of probability as a relative frequency.


***2. The following data from the US Census Bureau (http://www.census.gov/prod/2002pubs/p20-542.pdf) indicates the age distribution of voters in the 2000 presidential election:

Ages of voters Frequency (millions)
18 to 24 years old 32.3
25 to 44 years old 49.8
45 to 64 years old 64.1
65 years and over 67.6

a. What is the probability that a voter chosen randomly from these voters is 18 to 24 years old?

32.3/(32.3+49.8+64.1+67.6) ≈ 0.151

b. What is the probability that a voter chosen randomly from these voters is over 45 years old?

(64.1+67.6)/(32.3+49.8+64.1+67.6) ≈ 0.616

c.  Suppose you choose two random voters.  What is the probability that both are 18 to 24 years old?

Choosing two random voters can be viewed as essentially independent events, since the number of voters is so large.  (In general, choosing without replacement is not independent.)  So the probability is 0.151*0.151 ≈ 0.0228
 

***3. Suppose a family has four children.

a. Describe the sample space for the gender distribution (boy, girl, boy, boy etc) of the four children.  Note: the order matters!  There are sixteen possibilities.

BBBB, BBBG, BBGB, BBGG, BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG

b. If the probability of having a male baby is 0.512, what is the probability of having four girls?

The probability of having one girl is 1 – 0.512 = 0.488.  The probability of four girls is 0.488^4 ≈ 0.0567.  This event is fairly uncommon.

(skip) c. Simulate this scenario in Excel as follows.  In Column H type:

0 0.512
1 0.488

Then generate 10000 rows of  four 0's and 1's based on the theoretical probability.

0

Paste the top 10 rows of your simulation into your Word document.

Here is what I got. The fifth column as is the sum of the four columns.

0 0 1 0 1
0 1 1 1 3
0 0 0 0 0
1 0 0 0 1
1 1 1 0 3
1 0 1 1 3
0 1 0 0 1
1 1 1 1 4
1 0 0 0 1
0 0 0 0 0

(skip) d. Using your simulation, calculate the relative frequency of having all girls. You should get an answer fairly close to the answer you got in part b. (Hint.  The way I would do this is to add up each row and then count how many rows add up to 4, but there are other ways as well.)

I got 564/10000 = 0.0564 which is quite close to the theoretical 0.0567.

(skip) e. What is more likely in a family of four:  having two boys and two girls   or having three of one gender and one of the other gender? Your answer should include the calculation of the probability for each scenario.  Hint.  You can do this theoretically or you can use your simulation to  obtain a very close approximation to the theoretical probability. If you proceed theoretically, use the sample space to count how many possibilities there are in each case.  You have to consider two subcases in calculating the probability of having three of one gender and one of the other gender, namely the case of three girls and one boy  and the case of three boys and one girl. If you use your simulation, use COUNTIF.

The theoretical probability of having two boys and two girls is  6*(0.512)^2*(0.488)^2 ≈ 0.375  (there are 6 ways: BBGG, BGBG, BGGB, GBBG, GBGB, GGBB). The theoretical probability of having three of one and one of the other is   4*(0.512)^3*0.488 + 4*0.512*(0.488)^3 ≈ 0.5.

(skip) f. Did the result of part e surprise you? Why or why not?

This result surprised me the first time I heard it.  If you make a graph of the probabilities, one gets an idea of what is going on.  The two bars next to the middle bar add up to more than the middle bar. Intuitively, it seems that the two bars next to the middle one should be smaller than they really are.

***4.  The distribution of  ABO and Rh blood type for both men and women in the US is given in the table below:

   

ABO blood type
    O A B AB Total
Rh Factor Rh positive 38.4% 32.3% 9.4% 3.2%
Rh negative 7.7% 6.5% 1.7% 0.7%
 Total


The first thing I would do is calculate the marginal probabilities:

  ABO blood type  
    O A B AB Total
Rh Factor Rh positive 38.4% 32.3% 9.4% 3.2% 83.3%
Rh negative 7.7% 6.5% 1.7% 0.7% 16.6%
  Total 46.1% 38.8% 11.1% 3.9% 99.9%

Note that because of rounding, the total is 99.9% rather than 100%.

a. What is the probability that a randomly selected person is B negative?

1.7%

b. What is the probability that a randomly selected person in the US is Rh negative?

16.6%.

c. An individual with type B blood can safely receive transfusions only from persons with type B or type O blood.  What is the probability that the husband of a woman with type B blood is an acceptable blood donor for her?

46.1% + 11.1% = 57.2%

d. What is the probability that in a randomly chosen couple, the wife has type B blood and the husband has type A?

0.111*0.388 = 0.043

e. What is the probability that in a randomly chosen couple, one of the pair has type A blood and the other has type B?

2*0.111*0.388 = 0.086

(skip) f.  What is the probability that at least one of a randomly chosen couple has type O blood?

The easiest way to calculate this is to calculate the probability of the complementary event.  The complementary event is both not having O blood.  The probability of not having O blood is  1 – 0.461 = 0.539.  The probability of both not having O blood is 0.539*0.539.  So the probability of at least one person having O blood is 1 –0.539*0.539 = 0.709.  It is rather high likelihood.

(skip) g. There are medical risks for subsequent pregnancies if a mother and her child have different Rh factors.  What is the probability that a randomly chosen couple have different Rh factors?  (Hint. Remember that the male can be positive and the female negative and vice versa.)   (Biology note: It is possible for a child and mother to have different Rh factors even if they both parents have the same Rh factor if the parents are carriers of the negative Rh factor gene.)

The husband can be + and the wife can be – and vice versa.  So the probability is 2*(0.833)*(0.166) ≈ 0.277.

(skip) 5.  When tossing a coin, a "run" is a string of consecutive results that are all the same.  For example, in the sequence T T H H T T T T H T    there is a run of length 2 (two T's) followed by a run of length 2 (two H's), followed by a run of length 4 (four T's), followed by a run of length 1 (one H), and final run of length 1 (one T).  The longest run in this case is one of length 4.   Open the file SequencesOfTenCoinFlips.xls.  This file contains the results of 50,000 sequences of 10 coin flips. As usual,  "1" denotes heads, and "0" denotes tails.  In addition, all but the top ten rows of Column L are filled with the length of the longest run for the coin flip sequence in that row.

a. Fill in the top 10 rows and paste the top 10 rows of Column L into your Word document. (Hint. I'll get you started.  In row 1, the run lengths are 3, 1, 3, and 3, so the first number is 3.)

0 0 0 1 0 0 0 1 1 1   3
1 0 1 0 0 0 1 0 1 1   3
0 1 1 1 0 1 0 0 1 1   3
1 0 0 1 0 1 1 1 1 0   4
0 1 1 0 1 1 0 1 1 0   2
0 1 1 0 0 1 1 1 0 0   3
0 0 1 0 1 1 1 1 1 0   5
0 1 1 0 0 0 1 1 1 1   4
1 1 0 1 0 0 1 0 0 1   2
0 0 1 0 0 0 1 1 0 1   3

b. Fill in the following table based on the simulation data and paste it into your Word document.

 

Length of longest streak Frequency Percentage
1 92 0.2%
2 8397 16.8%
3 18155 36.3%
4 12454 24.9%
5 6193 12.4%
6 2745 5.5%
7 1161 2.3%
8 478 1.0%
9 225 0.5%
10 100 0.2%

Hint. The answer for example for 6 is:  6 was the longest streak in 2745 of the 50000 cases. So the frequency is 2745, and the percentage is 2745/50000 or about 0.0549.

c.  Based on the data in the table you created in part b, what is the probability that there will be a streak of 5 or more in a sequence of 10 coin tosses?

12.4% + 5.5% + 2.3% + 1.0% + 0.5% + 0.2% = 21.8%

d. Do you find the result in part c surprising? Why or why not?

The first time I came across this result, I was surprised.  Long streaks happen rather frequently!  One really shouldn't be too surprised to encounter a streak of 5 or more.

***7. What is the probability of getting a sum of 8 on a roll of two dice?

The sample space can be visualized as:

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

The cells shaded in yellow show how many ways one can get a sum of  8. The probability is 5/36 or about 0.139.

*** 8.  A common data collection method in consumer and market research is the telephone survey.  However, a major problem with consumer telephone surveys is nonresponse.  A nonresponse can result from a person not being home, a person not answering, a person refusing to participate, etc. Research has shown that the probability of a successful response to a telephone survey is 0.08.

a. What is the probability of a nonresponse on a random survey telephone call?

1 – 0.08 = 0.92.

b.  What is the probability of having at least one successful response out of three calls?

Once again, it easier to calculate the probability of the complementary event. The complementary event is the all three calls will be unsuccessful.  The probability that all three will be unsuccessful is 0.92^3 ≈ 0.779.  So the probability that at least one will be successful is 1 – 0.779 = 0.221.

c.  What is the probability that in a sequence of random telephone calls for a survey, that the first successful response will be on the fifth one?

0.92*0.92*0.92*0.92*0.08 ≈ 0.0573

(skip) d. (Extra credit)  What is the probability that it will take at least five phone calls before you have the first successful survey?

An elementary way calculating this probability is to make a table:

  Probability  
First success on first 0.08 0.0800
First success on second 0.92*0.08 0.0736
First success on third 0.92*0.92*0.08 0.0677
First success on fourth 0.92*0.92*0.92*0.08 0.0623
First success on fifth 0.92*0.92*0.92*0.92*0.08 0.0573
etc etc  

We want to calculate the part of the table that is shaded yellow, which is actually an infinite list.  Adding up the infinite list is possible, but it is easier to realize that the entire table has to add up to 1, so the probability we are interested in is 1 – (0.080 + 0.0736 + 0.0677 + 0.0623) ≈ 0.716.   So it would be extremely common to have to make at least 5 calls.