b) All outcomes are equally likely with probability = 1/6. The probabilities do add up to 1, as they must.

(1)

10. a. if you roll a single die and count the number of dots on top, what is the sample space of all possible outcomes?

b. assign probabilities to the outcomes of the sample space of part (a). Do the possibilities add up to 1? should they add up to 1?

a) {1, 2, 3, 4, 5, 6}

10. you roll two fair dice, a green one and a red one. b. find P(1 on green die and 2 on red die).

P = (1/6)2 = 1/36

c find P92 on green die and 1 on red die). P = (1/6)2 = 1/36

d. Find P(( 1 on green die and 2 on red die). P = (1/6)2 = 1/36

5.1

10. 25% of all state prison inmates released on parole become repeat offenders while on parole. Suppose the parole board is examining five prisoners up for parole. Let x = number of prisoners out of five on parole who become repeat offenders.

x: 0 1 2 3 4 5

P(x): 0.237 0.396 0.264 0.088 0.015 0.001

a. find the probability that one or more of the five parolees will be repeat offenders. How does this number relate to the probability that none of the parolees will be repeat

offenders?

P(x≥1)=0.396 + 0.264 + 0.088 + 0.015 + 0.001=0.764

P(x≥1)=1−P(0)=1−0.237=0.763

The small discrepancy is likely the result of roundoff error.

b. find the probability that two or more of the five parolees will be repeat offenders.

P(x≥2)=0.264 + 0.088 + 0.015 + 0.001 = 0.468

c. find the probability that four or more of the five parolees will be repeat offenders.

P(x≥4)=0.015 + 0.001 = 0.0.016

d. compute the mean, the expected number of repeat offenders out of 5. µ = x = xP(x)

x=0 5

(2)

e. compute the standard deviation of the number of repeat offenders out of five. σ = x2 − x 2 = x2P(x)

x=0 5

∑

−µ2 ₌ _0.969 ₌_0.984

14. sara is a 60 year old anglo female and is reasonably good health. she wants to take out a $50,000 term ( that is, straight death benefit) life insurance policy until she is 65. The policy will expire on her 65th birthday. the probability of death is given year is provided by the Vital Statistics section.

x=age: 60 61 62 63 64

P(death at this age):0.00756 0.00825 0.00896 0.00965 0.01035

a. Sara is applying to big rock insurance company for her term policy. what is the

probability that sara will die in her 60th year? Using the probability and the 50,000 death benefit, what is the expected cost to big rock insurance?

P(60) = 0.00756

E(C)=($50000)(0.00756)=$378

b. repeat part (a) for years 61, 62, 63, and 64. What would be the total expected cost to big rock insurance over the years 60 through 64?

age probability of death expected cost 60 0.00756 378.0 61 0.00825 412.5 62 0.00896 448.0 63 0.00965 482.5 64 0.01035 517.5 total 0.04477 2238.5

c. If big rock insurance wants to make a profit of $ 700 above the expected total cost paid out for sara's death, how much should it charge for the policy.

$2938.5 or $48.98/month.

d. If big rock insurance company charges $5000 for the policy, how much profit does the company expect to make?

$2761.50 5.2

12.What makes up a trial? An single observation of a random event or process. what is a success? A trial where the prescribed outcome is realized.

(3)

What are the values of, n, p and q? n is number of trials, p is the probability of a successful trial, and q = 1 – p is the probability of an unsuccessful trial.

Have you ever purchased an article of clothing worn it once and then returned the purchase? this is called a ont time fling. about 10% of all adults deliberately do a one time fling. In a group of seven adult friends what is that probability that:

a. no one has done a one-time fling? P(0)=(0.9)7 =0.478

b. at least one person has done a one-time fling?

P(x≥1)=1−P(0)=0.522

c. no more than two people have done a one-time fling? Let p=0.1 be the probability a person has a 1-time-fling

P(x≤2)= P(0)+P(1)+P(2) = 7! 7!p 0₍₁₋ _p₎7 ₊ 7! 6!1!p(1− p) 6₊ 7! 5!2!p(1− p) 2 ₌_0.974

What makes up a trial? what is a success? What is failure? What are the values of, n, p and q?

A trial is a random sampling of one a person. Success here is the outcome that the person has had a one-time-fling. Failure is the outcome that the person has not had a fling. There are 7 trials. The probability of success is 0.1, and the probability of failure is 0.9. 16. What makes up a trial? what is a success? What is failure? What are the values of, n, p and q? Trevor is interested in purchasing a sporting goods store. After examining his accounting records for the past several years, he found that the store has been grossing over $850 per day and about 60% of the business days it is open. Estimate that the probability that the store will gross over $850.

a. at least 3 out of 5 business days.

p=0.6 P(x≥3)= P(3)+P(4)+P(5) = 5! 2!3!(0.6) 3_(0.4)2₊ 5! 1!4!(0.6) 4_(0.4)1₊ 5! 0!5!(0.6) 5_(0.4)0 ₌_0.683

b. at least 6 out of 10 business days.

P(x≥6)=0.633

c. fewer than 5 out of 10 business days.

P(x≤5)=0.834

(4)

P(x<6)=0.0016

If this actually happened, might it shake you confidence in the statement p=0.60? Might it make you suspect that p is less than 0.60? Explain.

Yes. This is an exceedingly rare occurrence, and unless I knew from other means that p = 0.6, it would lead me to believe that p is indeed smaller than 0.6.

e. more than 17 out of the next 20 business days.

P(x>17)=0.003611

If this actually happened, might you suspect that p is greater than 0.60? explain. Yes, this is also a very rare occurrence, and would indicate p > 0.6.

5.3

10. The mountain states office of state farm insurance company reports that

approximately 85% of all automobile damage liability claims were made by people under 25 years of age. A random sample of five automobile insurance claims is under study. a. make a histogram showing the probability that r=0 to 5 claims are made by people under 25 years of age?

b. find the mean and standard deviation of this probability distribution. µ = Np=4.25 σ = npq =0.798

for sample size of 5, what is the expected number of claims made by people under 25 years of age?

µ = Np=4.25

22. It is reported that approximately 45% of all university professors are extroverted, suppose you have classes with six different professors.

a. what is the probability that all six are extroverts? P(6)=(0.45)6 =0.008304 0 0.1 0.2 0.3 0.4 0.5 0 1 2 3 4 5 pr obability (out of 5 r andom selections) no. people under 25

(5)

b. What is the probability that non of your professors is an extrovert? P(0)=(0.55)6 =0.027681

c. What is the probability that ar least two of your professors are extroverts?

P(x≥2)=0.836433

d. In a group of six professors selected at random, what is the expected number of extroverts? What is the standard deviation of the distribution?

µ = Np=2.7 σ = Npq =1.219

e. quota problem: ???