isds chapter 4

What is the probability that A^c and B4 occur?

0.12

Are events A and B independent?

No because P(A | B) ≠ P(A)

A sample space S yields five equally likely events, A, B, C, D, and E. a. Find P(D). (Round your answer to 2 decimal places.)

P(D)=0.20 explanation: P(D)= 1/5=.20

Find P(A U C U E)

.60 explanation: P(A U C U E) = P(A) + P(C) + P(E)= 1/5+1/5+1/5=3/5=.60

Let P(A) = 0.50, P(B) = 0.45, and P(A ∩ B) = 0.31.a. Are A and B independent events?

No because P(A | B) ≠ P(A)

B1 B2 B3. B4 A .09 .22 .15 .20 A^c .03 .10 .09 .12 Consider the following joint probability table. What is the probability that A occurs?

.09 +.22 +.15+.20=.66

What is the probability that B2 occurs?

.32 .22+.10=.32

Let P(A) = 0.57, P(B) = 0.22, and P(A | B) = 0.37. Calculate P(A ∩ B).

P(A ∩ B)=0.081 explanation: P(A ∩ B) = P(A | B)P(B) = (0.37)(0.22) = 0.081

A political reporter announces that there is a 40% chance that the next person to come out of the conference room will be a Republican, since there are 60 Republicans and 90 Democrats in the room.

Empirical probability

Given that A has occurred, what is the probability that B4 occurs?

0.3030 (+/-0.0001) P(B4 | A)= P(A ∩ B4)/P(A)= 0.20/0.66= 0.3030

Before flipping a fair coin, Sunil assesses that he has a 50% chance of obtaining tails.

Classical probability

At the beginning of the semester, John believes he has a 90% chance of receiving straight A's.

Subjective probability

An analyst estimates that the probability of default on a seven-year AA-rated bond is 0.41, while that on a seven-year A-rated bond is 0.59. The probability that they will both default is 0.10. What is the probability that at least one of the bonds defaults?

.90 P(A U B) = P(A) + P(B) − P(A ∩ B) = 0.41 + 0.59 − 0.10 = 0.90

Are events A and B mutually exclusive?

No because P(A ∩ B) ≠ 0.

Find P(B^c)

.80 explanation: First define B^c={A,C,D,E} P(B^c)=4/5=.80

Let P(A) = 0.50, P(B) = 0.45, and P(A ∩ B) = 0.31.a. Are A and B mutually exclusive events?

No because P(A ∩ B) ≠ 0.

Let P(A) = 0.57, P(B) = 0.22, and P(A | B) = 0.37. Calculate P(A U B).

P(A U B)=0.709 explanation: P(A U B) = P(A) + P(B) − P(A ∩ B) = 0.57 + 0.22 − 0.081 = 0.709

Jane Peterson has taken Amtrak to travel from New York to Washington, DC, on fourteen occasions, of which seven times the train was late. Therefore, Jane tells her friends that the probability that this train will arrive on time is 0.50. a.Would you label this probability as empirical or classical?

Empirical probability explanation: This is an empirical probability, since it is based on observed outcomes of an experiment.

An analyst estimates that the probability of default on a seven-year AA-rated bond is 0.41, while that on a seven-year A-rated bond is 0.59. The probability that they will both default is 0.10. What is the probability that neither the seven-year AA-rated bond nor the seven-year A-rated bond defaults? (Round your answer to 2 decimal places.)

.10 P((A U B)c) = 1 − P(A U B) = 1 − 0.90 = 0.10

What is the probability that a household is without kids and does not own an Apple product?

.1647 P(A^c ∩ B^c) = P(A^c) − P(A^c ∩ B) = P(A^c) − P(B | A^c)P(A^c) = 0.3167 − 0.48(0.3167) = 0.1647

An analyst estimates that the probability of default on a seven-year AA-rated bond is 0.41, while that on a seven-year A-rated bond is 0.59. The probability that they will both default is 0.10. Given that the seven-year AA-rated bond defaults, what is the probability that the seven-year A-rated bond also defaults?

.24 P(B | A) = P(A ∩ B)/P(A) = 0.10/0.41 = 0.24

Apple products have become a household name in America with 51% of all households owning at least one Apple product (CNN, March 19, 2012). The likelihood of owning an Apple product is 61% for households with kids and 48% for households without kids. Suppose there are 1,200 households in a representative community, of which 820 are with kids and the rest are without kids. Let event A correspond to "Household with kids", and B to "Household owns an Apple product". What is the probability that a household is without kids?

.3167 P(A^c)=1200-820/1200=380/1200=0.3167

What is the probability that neither A nor B takes place? (Round your answer to 2 decimal places.)

.36 P((A U B)c) = 1 − P(A U B);P(A U B) = P(A) + P(B) − P(A ∩ B) = 0.50 + 0.45 − 0.31 = 0.64;Therefore, P((A U B)c) = 1 − 0.64 = 0.36.

Apple products have become a household name in America with 51% of all households owning at least one Apple product (CNN, March 19, 2012). The likelihood of owning an Apple product is 61% for households with kids and 48% for households without kids. Suppose there are 1,200 households in a representative community, of which 820 are with kids and the rest are without kids. Let event A correspond to "Household with kids", and B to "Household owns an Apple product". What is the probability that a household is with kids and owns an Apple product?

.4168 P(A ∩ B) = P(B | A) P(A) = (0.61)(0.6833) = 0.4168

Given that B2 has occurred, what is the probability that A occurs?

.6875 P(A|B2) = P(A ∩ B2)/P(B2) = 0.22/0.32 = = 0.6875

What is the probability that A or B3 occurs?

.75 P(A U B3) = P(A) + P(B3) − P(A ∩ B3) = 0.66 + 0.24 − 0.15 = 0.75

The probabilities that stock A will rise in price is 0.61 and that stock B will rise in price is 0.39. Further, if stock B rises in price, the probability that stock A will also rise in price is 0.59. What is the probability that at least one of the stocks will rise in price?

.77 P(A U B) = P(A) + P(B) − P(A ∩ B) = P(A) + P(B) − P(A | B)P(B) = 0.61 + 0.39 − 0.59(0.39) = 0.77

Let P(A) = 0.57, P(B) = 0.22, and P(A | B) = 0.37. Calculate P(B | A).

P(B | A)= 0.142 explanation: P(B | A) =P(A ∩ B)/P(A) = .081/.57 = .142

Jane Peterson has taken Amtrak to travel from New York to Washington, DC, on fourteen occasions, of which seven times the train was late. Therefore, Jane tells her friends that the probability that this train will arrive on time is 0.50. b. Why would this probability not be accurate?

The experiment must be repeated a large number of times for empirical probabilities to be accurate. explanation: This probability is unlikely to be accurate due to the small sample.

Apple products have become a household name in America with 51% of all households owning at least one Apple product (CNN, March 19, 2012). The likelihood of owning an Apple product is 61% for households with kids and 48% for households without kids. Suppose there are 1,200 households in a representative community, of which 820 are with kids and the rest are without kids. Let event A correspond to "Household with kids", and B to "Household owns an Apple product". Are the events "household with kids" and "household without kids" mutually exclusive and exhaustive?

Yes P(A ∩ A^c) = 0. Events corresponding to "household with kids" and "household without kids" are mutually exclusive. They are also exhaustive because P(A) + P(A^c) = 1.