practice questions..

[Akshay Jain]

1.Suppose that two persons make an appointment to meet between 5 p.m. and 6 p.m. at a certain location, and they agree that neither person will wait more than 10 minutes for the other person. If they arrive independently at random times between 5 p.m. and 6 p.m., what is the probability that they will meet?

a)     7/16

b)    9/16

c)     11/36      

d)    25/16

2. A candy factory has an endless supply of red, orange, yellow, green, blue, black, white, and violet jellybeans. The factory packages the jelly beans into jars in such a way that eachjarhas200 beans, equal number of red and orange beans, equal number of yellow and green beans, one more black bean than the number blue beans, and three more violet beans than the number of white beans. One possible color distribution, for example, is ajar of 50 yellow,50 green, one black, 48 white, and 51 violet jelly beans. As a marketing gimmick, the factory guarantees that no two jars have the same color distribution. What is the maximum number of jars the factory can produce?

a)       200C3

b)       98C3

c)       102C3

d)       101C3

3. Alice, Bob, and Caroll play a chess tournament. The first game is played between Alice and Bob. The player who sits  out a given game plays next the winner of that game. The tournament ends when some player wins two successive games. Let a tournament history be the list of game winners, so for example ACBAA corresponds to the tournament where Alice won games 1, 4, and 5, Caroll won game 2, and Bob won game 3.

(a)    We are told that every possible tournament history that consists of k games has probability 1/2k, and that a tournament history consisting of an infinite number of games has zero probability. Assuming the probability law to be correct,

find the probability that the tournament lasts no more than 5 games,

a)       5/16

b)      15/16

c)       1/16

d)      7/16
 

 The probability for each of Alice, Bob, and Caroll winning the tournament.

a)       5/14, 5/14, 4/14

b)      4/14, 4/14, 6/14

c)       4/14, 5/14, 5/14

d)      5/14, 4/14, 5/14

4.A retired couple have a dog, and are trying to decide how to allocat their time. Assume that partner j (j=l ,2) can allocate her/his time between two activities: leisure, or walking the dog (hj). Each partner has a total of T hours of time available. Total hours dog-care are h=hj+h2. Assume that partner #1 has a utility function of the form

U1(l1,h) = βlog(l1) + (1-β)log(h)

while partner #2 has a utility function:

U2(l2,h) = βlog(l2) + (1-β)log(h)

Note that both partners value leisure and the total amount of time devoted to the dog (h).

Suppose that partners make their time choices taking each other's choices as given. Find l's optimal choice for h1,taking h2 as given, and 2's optimal choice for h2, taking h1 as given. Solve   for the total time spend walking the dog in the Cournot equilibrium, hc.

a)            T(1-β)/2(1+β)

b)           T(1-β)/(1+β)

c)           2T(1-β)/(1+β)

d)           None of these

I havnt tried these questions so dont no the answers...but u pple can discuss.....

[mrittik]

1st one is 11/36...

[jane]

how are you getting this answer?

[mrittik]

very easy....2*integrate to 0 to 5/6 of xdx.....1-the integration result

[jane]

m sorry, but i didn't get it... can u elaborate a bit? i'm kinda dumb

[mrittik]

only 10 minutes to meet with each other.....so the the rest of the time is 1hr-10/60hr=5/6hr...so to get the probability it is 1-P(they cant meet)=1-as state above

[Dreyfus]

For q4..I m getting h1*=h2*=T(1-B)/(1+B)
So that h=2T(1-B/(1+B)