ECSE-2500, Engineering Probability, Spring 2010, Rensselaer Polytechnic Institute

Exam 1, 2pm Tues March 2 2010

You may write free as the correct solution to any 2 5-point question parts. That means that answering 100 points is a complete exam.

1. Two teams are playing a 5 game series. Assume that they are equally matched. What is the probability that
1. (5 points) the series winner will be known after 3 games (i.e., the same team wins the 1st 3 games)? 1/4
2. (5) the series winner will be known after exactly 4 games? 3/8
3. (5) all five games must be played to determine the winner? 3/8
2. This question is about repeatedly tossing a fair die with 6 sides.
1. (5) What is the probability that the first toss is a 1? 1/6
2. (5) What is the probability that the 2nd toss is a 1, regardless of the 1st toss? 1/6
3. (5) Are the above two events independent? No need to explain, just say yes or no. yes, the tosses don't affect each other.
4. (5) What is the probability that the 2nd toss is a 1, and this is the first time you got a 1? 5/36
5. (5) What is the probability that the k-th toss is the first 1? (5/6)^(k-1) (1/6)
3. For this question, the sample space S is real numbers from 0 to 1. Let A be the event that {$1/4\le x\le 1/2$}, and B be the event that {$1/3\le x\le 2/3$},
1. (5) What is the probability of A? 1/4
2. (5) What is the probability of B? 1/3
3. (5) What is the probability of {$A\cap B$} ? 1/6
4. (5) What is P[A|B]? 1/2
5. (5) Are A and B independent? Why? no since 1/4 * 1/3 != 1/6
4. There are 4 applicants for a 2 person team.
1. (5) How many different teams can I form? 4 choose 2 = 6
2. (5) Two of the applicants are Able and Baker. What's the probability that both of them are on the team (assuming picks are random)? 1/6
3. (5) What is the probability that Able is on the team, given that Baker is on the team?
first need P(B) = (3 choose 2)/(4 choose 2) = 1/2
P(A|B) = P(AB)/P(B) = 1/6 / (1/2) = 1/3
4. (5) Are the events of Able being chosen and Baker being chosen independent? Why?
No. P(AB) = 1/6 != P(A) P(B) = 1/4. Informally, choosing Able makes it less likely that Baker will be chosen.
5. There are 4 guides and 3 tour groups that need guiding, each on a different day. That is, there is no conflict with having one guide handling more than 1, or even all 3, groups.
1. (5) How many different ways can guides be assigned to tour groups? 4^3 = 64
2. (5) One guide is named Able. What's the probability that Able is never picked (assuming randomness etc)?
(3/4)^3 = 27/64
3. (5) What's the probability that Able is picked exactly twice (assuming randomness etc)?
(3 choose 2) * 3 / 64 = 9/64
6. I buy widgets from Acme 30% of the time and from Supreme 70%. 20% of Acme widgets are bad, and 10% of Supreme widgets are bad.
1. (5) What percentage of the widgets that I buy are bad?
.3*.2+.7*.1= .13 = 13%
2. (5) Given I have a bad widget, what's the probability that it's from Acme?