Midterm 2: Chapter 4 Multiple Choice

A college plans to interview 8 students for possible offer of graduate assistantships. The college has three assistantships available. How many groups of three can the college select?

(8) = 8! / (8-3)! 3! = 56 of three can be selected (3)

All the employees of ABC Company are assigned ID numbers. The ID number consists of the first letter of an employee's last name, followed by four numbers. A.How many possible different ID numbers are there? B.How many possible different ID numbers are there for employees whose last name starts with an A?

A) the name can start with 26 letters and four numbers can be 10000 ways so total 26*10000 = 260000types possible B)for employees with A there are 10000 different IDs

If A and B are independent events with P(A) = 0.4 and P(B) = 0.25, then P(AÈB) = A.0.65 B.0.55 C.0.10 D.0.75

ANS: B

If P(A) = 0.58, P(B) = 0.44, and P(AÇB) = 0.25, then P(AÈB) = A.1.02 B.0.77 C.0.11 D.0.39

ANS: B

A graphical device used for enumerating sample points in a multiple-step experiment is a A.bar chart B.pie chart C.histogram D. None of these alternatives is correct.

ANS: D

A sample point refers to the A.numerical measure of the likelihood of the occurrence of an event B.set of all possible experimental outcomes C.individual outcome of an experiment D.sample space

C

An experiment consists of four outcomes with P(E1) = 0.2, P(E2) = 0.3, and P(E3) = 0.4. The probability of outcome E4 is A.0.500 B.0.024 C.0.100 D.0.900

C

If P(A) = 0.50, P(B) = 0.60, and P(AÇB) = 0.30, then events A and B are A.mutually exclusive events B.not independent events C.independent events D.not enough information is given to answer this question

C

The addition law is potentially helpful when we are interested in computing the probability of A.independent events B.the intersection of two events C.the union of two events D.conditional events

C

A method of assigning probabilities based upon judgment is referred to as the A.relative method B.probability method C.classical method D.subjective method

D

Bayes' theorem is used to compute A.the prior probabilities B.the union of events C.intersection of events D.the posterior probabilities

D

If P(A) = 0.45, P(B) = 0.55, and P(AÈB) = 0.78, then P(A|B) = A.zero B.0.45 C.0.22 D.0.40

D

If X and Y are mutually exclusive events with P(X) = 0.295, P(Y) = 0.32, then P(X|Y) = A.0.0944 B.0.6150 C.1.0000 D.0.0000

D

From a group of six finalists to a contest, three individuals are to be selected for the winner awards. Determine the number of possible selections. From a group of six finalists to a contest, three individuals are to be selected for the first and second and third places. Determine the number of possible selections.

Ist place can be taken by 7 people, IInd place can be taken by 6 people and finally IIIrd place can be taken by 5 people: so the possible arrangements will be: 7 x 6 x 5= 210

Three workers at a fast food restaurant pack the take-out chicken dinners. John packs 45% of the dinners but fails to include a salt packet 4% of the time. Mary packs 25% of the dinners but omits the salt 2% of the time. Sue packs 30% of the dinners but fails to include the salt 3% of the time. You have purchased a dinner and there is no salt.Find the probability that Mary packed your dinner.

John-packed, salt-less dinners: 0.45 * 0.04 = .0180 Mary-packed, salt-less dinners: 0.25 * 0.02 = .0050 Sue- packed, salt-less dinners: 0.30 * 0.03 = .0090 The total proportion of salt-less dinners is .0180 + .0050 + .0090 = .0320 (3.2% of the dinners have no salt.) Mary's share of this amount is .0050 / .0320 = 5 / 32

Ten individuals are candidates for positions of president, vice president of an organization. How many possibilities of selections exist?

possibilities = 10C1 x 9C1 [ for each of president, vice president] = 90 possibilities.

A statistics professor has noted from past experience that a student who follows a program of studying two hours for each hour in class has a probability of 0.9 of getting a grade of C or better, while a student who does not follow a regular study program has a probability of 0.2 of getting a C or better. It is known that 70% of the students follow the study program. Find the probability that a student who has earned a C or better grade, followed the program.

student who follows a program of studying two hours for each hour in class has a probability of = 0.9 who doesnt follow : 0.2 known that 70% of the students follow the study program and 30 % doesn't so total probability of earning c or above = 0.9x0.7 + 0.3x0.2 =0.69 so the probability that a student who has earned a C or better grade, followed the program = = probabilty that student follow each class* no of % of student / total probability of earning c or abobe = 0.9*0.7/0.69 = 0.913 ANS.