2. Suppose a professor randomly selects three new teaching assistants from a total of 10 applicants, six male and four female. What is the probability that the professor hires no female students?
Hiring no female means hiring all 3 malesP(Hiring 3 male students) = Number of ways of selecting 3 males out of 6 / Number of ways of selecting 3 students out of 10 students
= C(6, 3)/C(10, 3)
= 0.1667
3. The following data are from a sample of 1,200 voters in last year's Presidential election: 260 McCain voters were 18-29 years of age; 390 McCain voters were 30 years and older; 385 Obama voters were 18-29 years of age; 165 Obama voters were 30 years and older. Please answer the questions below about this sample.
Age / McCain / Obama / Total18 - 29 / 260 / 385 / 645
> 30 / 390 / 165 / 555
Total / 650 / 550 / 1200
a. Given that a voter is in the 18-29 years group, what is the probability that he/she voted for McCain?
= 260/645 = 0.4031
b.Given that a voter is in the 30 and over group, what is the probability that he/she voted for McCain
= 390/555 = 0.7027
c. What is the probability that a McCain voter is 30 and over= 390/650 = 0.60
d. What is the probability that an Obama voter is 30 and over= 165/550 = 0.30
4. A recent preseason NCAA football poll asked respondents to answer the question, "Will the Big Ten or the Pac-10 have a team in this year's national championship game, the Rose Bowl?" Of the 13,429 respondents, 2,961 said the Big Ten would; 4,494 said the Pac-10 would; 6,823 said neither the Big Ten nor the Pac-10 would have a team in the Rose Bowl; and 849 said that both the Big Ten and the Pac-10 would have a team in the Rose Bowl.
a. What is the probability that a respondent said neither the Big Ten nor the Pac-10 would have a team in the Rose Bowl= 6823/13429 = 0.5081b. What is the probability that a respondent said either the Big Ten or the Pac- 10 or both would have a team in the Rose Bowl = 1 - probability that a respondent said neither the Big Ten nor the Pac-10 would have a team in the Rose Bowl = 1 - 0.5081 = 0.4919
c. What is the probability that a respondent said that both the Big Ten and the Pac-10 would have a team in the Rose Bowl= 849/13429 = 0.0632
5. Consider the situation of a clothing retailer. The owner knows from experience that the probability that any one customer will make a purchase is .30.
p = 0.3, q = 1 - 0.3 = 0.7, n = 3Binomial probability formula: P(x) = nCx p^x q^(n - x)
a. What is the probability that none of the next 3 customers entering the store will make a purchase?
= 3C0 0.3^0 0.7^3 = 0.343
b. What is the probability that exactly 1 of the next 3 customers entering the store will make a purchase
= 3C1 0.3^1 0.7^2 = 0.441
c. What is the probability that exactly 2 of the next 3 customers entering the store will make a purchase
= 3C2 0.3^2 0.7^1 = 0.189
d. What is the probability that exactly 3 of the next 3 customers entering the store will make a purchase
= 3C3 0.3^3 0.7^0 = 0.027
6. A manufacturer currently receives shipments of parts from a supplier....who ships both defective as well as non-defective parts in each shipment. Suppose a quality control inspector at the manufacturer randomly selects three of the twelve parts from a shipping container, and that (unknown to the inspector) the container has five defective parts and seven non-defective parts.
a. What is the probability of finding no defective parts?= number of ways of selecting 3 good parts out of 7 / number of ways of selecting 3 parts out of 12
= C(7, 3)/C(12, 3) = 0.1591
b. What is the probability of finding one defective part?
= (number of ways of selecting 1 defective part out of 5) * (number of ways of selecting 2 good parts out of 7) / number of ways of selecting 3 parts out of 12
= C(5, 1) * C(7, 2)/C(12, 3) = 0.4773
c. What is the probability of finding two defective parts?
= (number of ways of selecting 2 defective part out of 5) * (number of ways of selecting 1 good part out of 7) / number of ways of selecting 3 parts out of 12
= C(5, 2) * C(7, 1)/C(12, 3) = 0.3182
d. What is the probability of finding three defective parts?
= number of ways of selecting 3 defective parts out of 5 / number of ways of selecting 3 parts out of 12
= C(5, 3)/C(12, 3) = 0.0455
7. A police officer must score in the upper 2.5% of an exam to qualify for the rank of ‘detective.' If scores are normally distributed with a mean of 100 and a standard deviation of 15, what score must an officer have to qualify for the rank of ‘detective'?
Area under the standard normal curve is 2.5%, that is 0.025 to the right of z = 1.96x = μ + z * = 100 + 1.96 * 15 = 129.4
An officer should have a score of 129.4 and above.
8. Recall that physicians are now using Bayesian reasoning (including prior, conditional, joint and posterior probabilities) in the medical diagnostic phase of patient examinations. Whenever a patient has an undiagnosed disease, the examining physician initially develops prior probabilities of diseases the patient might have. Typically, these probabilities are assigned using the "subjective method," and are usually a function of such things as the physician's view of various relevant environmental factors such as the occurrence of epidemics, geography, recent weather patterns, and climate.
- Suppose a physician believes a patient has one of two possible diseases, denoted D1 and D2 with P(D1) = 0.30 and P(D2) = 0.70.
- Suppose also that medical research has established the probability associated with each of three symptoms (denoted S1, S2, and S3) that may accompany the two diseases. That is, suppose that, given diseases D1 and D2, the probabilities that the patient will have symptoms S1, S2, or S3 are as follows: P(S1|D1) = .25; P(S2|D1) = .15; P(S3|D1) = .65; P(S1|D2) = .10; P(S2|D2) = .15; P(S3|D2) = .20.
- After a certain symptom is found to be present, the medical diagnosis may be aided by finding the revised probabilities of each particular disease. Compute the posterior probabilities of each disease given the following medical findings.
a. The patient has symptom S1. D1 ______and D2 ______
b. The patient has symptom S2. D1 ______and D2 ______
c. The patient has symptom S3. D1 ______and D2 ______
d. For a patient with symptom S1 in part (a), we also find symptom S3. The posterior probabilities are: D1 ______and D2 ______
I am not sure I understand the notation S1 D1 ____, D2 ____ etc. Can you please elaborate and clarify?9. A purchasing agent has placed "rush orders" for a particular raw material with two different suppliers, A and B. If neither order arrives in 4 days, the production process must be shut down until at least one of the orders arrives. The probability that supplier A can deliver the material in 4 days is .50, while the probability that supplier B can deliver the material in 4 days is .40.
a. What is the probability that both suppliers will deliver the material in 4 days? = 0.5 * 0.4 = 0.20b. What is the probability that at least one supplier will deliver the material in 4 days?
= P(A supplies and B does not) + P(B supplies and A does not) + P(Both A and B supply)
= 0.5 * (1 – 0.4) + 0.4 * (1 – 0.5)+ 0.2 = 0.7
c. What is the probability that the production process will be shut down in 4 days because of a shortage of raw material?
= P(Neither A nor B will supply) = 1 - P(At least one of A and B will supply) = 1 - 0.7 = 0.3
10. In a study of the consumer's view of the economy, the probability thata consumer would buy a house during the year was .053, and the probability that a consumer would buy a car during the year was .198. There was also a .009 probability that a consumer would buy a house and a car during the year.
a. What is the probability that a consumer would buy either a car or a house during the year?= P(car) + P(house) - P(car and house) = 0.198 + 0.053 - 0.009 = 0.242
b. What is the probability that a consumer would buy a car during the year given that the consumer purchased a house?
P(car | house) = P(car and house)/P(house) = 0.009/0.053 = 0.1698
11. 50% of drivers arriving at the Holland Tunnel have correct change.
a. If 20 autos pass through the toll plaza in the next 3 minutes, what is the probability that between 5 and 15 cars, inclusive, will have the exact change?p = 0.5, q = 1 - p = 0.5, n = 20, x1 = 4.5, x2 = 15.5
μ = np = 10, = (npq) = 2.236
z1 = (x1 - μ)/ = (4.5 - 10)/2.236 = -2.4597 and z2 = (x2 - μ)/ = (15.5 - 10)/2.236 = 2.4597
P(5 ≤ x ≤ 15) = P(-2.4597 < z < 2.4597) = 0.9861
b. If 85% of autos arriving at the tunnel's toll booth have either New York or New Jersey license plates, what is the probability that (of these 20 autos) 10 or more (i.e., between 10 & 20 inclusive) will bear license plates from states other than New York or New Jersey?
q = 0.85, p = 1 – q = 0.15, n = 20, x = 9.5
μ = np = 3, = (npq) = 1.5969
z = (x - μ)/ = (9.5 - 3)/1.5969 = 4.0705
P(x 10) = P(z 4.0705) = 0
12. People arrive at a customer service desk at the rate of 45 per hour.
(a) What is the probability that two or fewer people will arrive in the next 6 minutes?m = (45/60) * 6 = 4.5, x = 2
Poisson probability formula is P(x) = e^-m * m^x / x!
P(x ≤ 2) = P(0) + P(1) + P(2)
= e^-4.5 * 4.5^0 / 0! + e^-4.5 * 4.5^1 / 1! + e^-4.5 * 4.5^2 / 2! = 0.1736
(b) What is the probability that the customer service representative can take a two-minute coffee break andarrive back at the desk finding no one waiting to be served?
m = (45/60) * 2 = 1.5, x = 0
Poisson probability formula is P(x) = e^-m * m^x / x!
P(0) = e^-1.5 * 1.5^0 / 0! = 0.2231
- Suppose that a hand of 12 cards is dealt from a normal deck of 52 playing cards. What is the probability that of the 12 cards, 3 will be diamonds, 3 will be clubs, 5 will be spades, and 1 will be hearts?
Since there are 13 cards of each type, the probability is C(13, 3) * C(13, 3) * C(13, 5) * C(13, 1) / C(52, 12)
= 0.0066
- What is the probability that in a group of 25 randomly selected people two or more will be found to share the same birthday? Please ignore leap years (like 2008)
P(At least two people out of n share a birthday)
= 1 - P(No two people out of n share a birthday)
= 1 - P(Everyone of the n people has a different birthday)
= 1 - [(365 * 364 * 363 * ... * (365 - n + 1)/(365^n)]
When n = 25,
P(At least two people out of 25 share a birthday) = 1 - [(365 * 364 * ... * 341)/365^25] = 1 – 0.4313 = 0.5687
- Suppose that you are given the following information: (a) in random testing, you test positive for a disease; (b) all people who have the disease test positive for it; (c) in 5 percent of cases, this test shows positive even when the subject does not have the disease; and, (d) in the population at large, one person in 1,000 has the disease. What is the probability that you have the disease?
P(you have the disease | you test positive) = P(you test positive | you have the disease) * P(you have the disease) / P(you test positive)
= 1 * 0.001 / {1 * 0.001 + 0.05 * (1 - 0.001)}
= 0.0196.