Test 1_ Winter 2007

STAT 505 TEST 1

NAME___________________________

FOR FULL CREDIT CLEARLY EXPLAIN YOUR REASONING. SIMPLY WRITING YOUR ANSWER, WITHOUT ANY EXPLANATION, ONLY ACOUNTS FOR HALF POINTS.

a. A= Yankees, B=Ac= Mets E=employed, U=Ec=unemployed P(A)= 0.4, P(B) = 0.6, P(E/A)=0.81, P(E/B)=0.95 P(U) = P(U/A) P(A) + P(U/B) P(B) = 0.19*0.40+0.05*0.60 =0.106

b. P(A/U) = P(U/A) * P(A) / P(U) = 0.19*0.40/00.106 = 0.717

2. 20 Points. A fleet of seven taxis is to be randomly assigned to three airports A, B, and C, with two going to A, four to B and one to C. a. (10 Points) In how many ways can this be done?

7! / (2! 4! ) = 5 6 7 / 2 = 105 ways

b. (10 Points) What is the probability that the specific cab driven by Mark is assigned to airport C?

6! / (2! 4!) = 15 ways P(A)= 15/105 =1/7

3. (15 Points) Marie is getting married tomorrow, at an outdoor ceremony in the desert. In the recent years, it has rained only 5 days each year. Unfortunatelly, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it does not rain, he incorrectly forecasts rain 10% of the time. What is the probability that it will rain on the day of Marie’s wedding?

A ={it rains at Marie’s wedding} B={It does not rain at marie’s wedding} C = {Weatherman predicts rain} P(A) = 5/365 = 0.01369 P(B) = 360/365 =0.98630 P(C/A) =0.90 P(C/B) = 0.10 P(A/C) = {P(C/A) *P(A) }/{P(C)}= {P(C/A) *P(A) }/{ P(C/A) *P(A) + P(C/B) *P(B) } =0.111 When the weatherman predicts rain, it actuallu rains only about 11% of the time.

4. (20 Points) A random sample of freshman at a large university shows the following: • 31% take History(Hist) • 40% take Chemistry(Chem) • 10% take Hist and Chem • 5% take Chem and Econ • 5% take Hist and Econ • 3% take all 3 • 30% take none. Draw the corect Venn-Diagram. – 5 points. Answer the following (2.5 points each) showing how you calculate : a. What proportion of freshman take Econ? b. What proportion of freshman take only History? c. What proportion take History and Chemistry, but not Econ. d. What proportion take at least one of the 3 courses? e. What proportion take all the 3 courses? f. What proportion take none of these courses? P(H ∩E)=0.05 P(C ∩ E) =0.05 P(C∩H)=0.10 P(only H) = 0.31-0.02-0.03-0.07=0.19 P(only C) = 0.40-0.02-0.07-0.03=0.28 P(only E) = 0.70-0.19-0.28-0.14=0.09

5. (20 points) Suppose that you are the chairman of a consortium of three groups of advisors: one group of 10 computer scientists, 8 educators, and 4 child behaviorists. You want to appoint some committees from among these 22 advisors. a. (5 Points) In how many ways can you appoint a five-member committee?

There are 22 objects and 5 boxes. are 26334 ways of appointing a five-member committee.

. There

b. (5 Points) In how many ways can you appoint a five-member committee with at least one computer scientist?

We know there are 26334 ways of appointing a 5-member committee, from (a). Now, we can find the number of ways of appointing a 5-member committee without a single computer scientist. Then we can subtract that number from 26334 to get the answer for this question. So, we choose 0 computer scientist, and then that leaves us with 12 people to form the 5-member committee. And there are ways. So, the number of ways that he can appoint a 5-member committee with at least 1 statistician is 26334 – 792 = 25542.

c. (5 Points) A certain computer scientist, Dr. G, and an educator, Dr. P, refuse to serve together on the same committee. In how many ways can you appoint a five-member committee that does not contain both Dr. G and Dr. P? First, let us decompose the problem into several scenarios: (1) where both Dr. G and Dr. P are not in the committee, (2) where Dr. G is in the committee, and (3) where Dr. P is in the committee. For the first scenario, there are 20 people left and we choose 5 out of it. Thus, there are ways. For the second scenario, if Dr. G is in the committee, then Dr. P cannot be in the committee. So, out of the four positions left, we have only 20 people to

choose from. So, there are ways. For the third scenario, since it is similar to the second scenario, we also have 4845 ways. Now, since all three scenarios are mutually exclusive, we can simply add them up: 15504 + 4845 + 4845 = 25194. There are 25194 ways of appointing a fivemember committee with the specified constraint. d. (5 Points) In how many ways that a five-member committee can be appointed such that the number of educators is greater than the number of computer scientists, and the number of computer scientists is greater than the number of child behaviorists?

We want to make sure that the number of educators is greater than the number of computer scientists, and the number of computer scientists is greater than the number of child behaviorists. First, let us see in how many ways that can happen. Suppose that [4 1 0] means that the number of educators is 4, the number of computer scientists is 1, and the number of child behaviorists is 0. Thus [4 1 0] is a valid committee under the constraint. [5 0 0] is not, for example. There is one other valid way: [3 2 0]. Now, for the [4 1 0] case, there are

ways of

selecting four educators out of 8, and there are ways of selecting 1 computer scientist out 10. Thus, there are a total of (70)(10) = 700 ways for the [4 1 0] case. Now for the [3 2 0] case, there are

ways of selecting

two computer scientists out of 10, and there are ways of selecting 2 computer scientists out of 8. Thus, there are a total of (45)(56) = 2520 ways for the [3 2 0] case. Since the two cases are mutually exclusive, we can simply add them up. Thus, we have a total of 700 + 2520 = 3220 ways of appointing a 5member committee under the above constraint.

FORMULAS 1. Probability of an event E: P(E) = {number of outcomes in the event} / { total number of outcomes in the sample space} 2. Theorem. The number of ordered arrangements, or permutations, of r objects selected from n distinct objects (r ≤ n) is given by . n! Prn = n(n − 1)...(n − r + 1) = (n − r )! 3. Theorem. The number of distinct unordered subsets, or combinations, of size r that can be selected from n distinct objects (r ≤ n) is given by n n!   =  r  r!(n − r )! 4. Theorem. The number of ways of partitioning n distinct objects into k groups containing n1, n2 … nk objects respectively, is n! n1! n2 !...nk ! where k

n = n ∑ i= 1

i

5. Definition: If A and B are any two events, then the conditional probability of A given B, denoted P(A|B), is P ( AB ) P( A | B) = P( B) provided that P(B) >0. 6. Definition: Two events are said to be independent if P(A|B)=P(A) or P(B|A)=P(B). This is equivalent to stating that P(AB)=P(A)P(B). 7. Theorem. If A and B are any two events, then P(A U B) = P(A) + P(B) – P(AB). If A and B are mutually exclusive, then P(A U B) = P(A) + P(B). 8. Theorem. If A and B are any two events, then P(A ∩ B) = P(A) P(B|A) = P(B)P(A| B). 9. Bayes’ Rule

P ( B j | A) =

P( B j ) P( A | B j ) k

∑ P( B ) P( A | B ) i =1

i

i