Test 2 Stats

f P(A | B) = 0.40 and P(B) = 0.30, find P(A∩B). Select one: A. .525 B. .120 C. .571 D. .171

B. .120

The number of unique orders in which five items (A, B, C, D, E) can be arranged is: Select one: A. 5. B. 120. C. 24. D. 84

B. 120

Compliment

P(A)+P(A')=1

Mutually Exclusive

(Disjoint) A and B happening at the same time = 0

Probability

1= Will occur 0= cannot occur

6C2

6!/2!(6-2!)=

The value of 6C2 is: Select one: A. 15. B. 720. C. 30. D. 12.

A. 15

Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent servers in parallel, each of which is 95 percent reliable. What is the smallest number of independent file servers that will accomplish the goal? Select one: A. 4 B. 3 C. 1 D. 2

B. 3

If two events are collectively exhaustive, what is the probability that one or the other will occur? Select one: A. 0.50 B. Can't tell from given information C. 1.00 D. 0.00

C. 1.00

The probability that event A occurs, given that event B has occurred, is an example of: Select one: A. more than one of the above. B. a marginal probability. C. a conditional probability. D. a joint probability.

C. A conditional Probability

At Dolon General Hospital, 30 percent of the patients have Medicare insurance (M) while 70 percent do not have Medicare insurance (M´). Twenty percent of the Medicare patients arrive by ambulance, compared with 10 percent of the non-Medicare patients. If a patient arrives by ambulance, what is the probability that the patient has Medicare insurance? Select one: A. .5000 B. .1300 C. .7000 D. .4615

D. .4615

P(AUB)

Union= A or B

Collectively Exhaustive

all events that can occur

If P(A∩B) = 0.50, can P(A) = 0.20? Select one: A. Only if P(B∩A) = 0.60 B. If P(A) = 0.20, then P(A∩B) cannot equal 0.50. C. Not unless P(B) = 0.30 D. Only if P(A | B) = 0.10

B. If P(A) = 0.20, then P(A∩B) cannot equal 0.50

In a certain city, 5 percent of all drivers have expired licenses, 10 percent have an unpaid parking ticket, and 1 percent have both an expired license and an unpaid parking ticket. Are these events independent? Select one: A. Can't tell from given information B. No C. Yes

B. No

Regarding probability, which of the following is correct? Select one: A. The probability of the union of two events can exceed one. B. When two events A and B are independent, the joint probability of the events can be found by multiplying the probabilities of the individual events. C. When events A and B are mutually exclusive, then P(A∩B) = P(A) + P(B). D. The union of events A and B consists of all outcomes in the sample space that are contained in both event A and event B.

B. When two events A and B are independent, the joint probability of the events can be found by multiplying the probabilities of the individual events.

Which statement is false? Select one: A. If P(A) = .05, then the odds against event A's occurrence are 19 to 1. B. The number of permutations of five things taken two at a time is 20. C. If A and B are mutually exclusive events, then P(A or B) = 0

C. If A and B are mutually exclusive events, then P(A or B) = 0.

Regarding the rules of probability, which of the following statements is correct? Select one: A. If A and B are independent events, then P(B) = P(A)P(B). B. The sum of two mutually exclusive events is one. C. If event A occurs, then its complement will also occur. D. The probability of A and its complement will sum to one.

D. The probability of A and its complement will sum to one.

Events A and B are mutually exclusive when: Select one: A. they are independent events. B. P(A)P(B) = P(A | B) C. P(A)P(B) = 0 D. their joint probability is zero.

D. their joint probability is zero.

Two events are complementary (i.e., they are complements) if: Select one: A. they are independent events with equal probabilities. B. the joint probability of the two events equals one. C. the sum of their probabilities equals one. D. they are disjoint and their probabilities sum to one.

D. they are disjoint and their probabilities sum to one.

P(A∩B)

Intersection (Joint) A and B

P(AUB)=

P(A)+P(B)-P(A∩B)

P(A∩B)=

P(A/B)P(B)