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

# Exam 1, 2pm Tues March 2 2010

**Name, RCSID, RIN:**
answers

*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.*

- Two teams are playing a 5 game series. Assume that
they are equally matched. What is the probability that
*(5 points)*the series winner will be known after 3 games (i.e., the same team wins the 1st 3 games)? 1/4*(5)*the series winner will be known after exactly 4 games? 3/8*(5)*all five games must be played to determine the winner? 3/8*(0)*Do your 3 answers total 1?

- This question is about repeatedly tossing a fair die with 6 sides.
*(5)*What is the probability that the first toss is a 1? 1/6*(5)*What is the probability that the 2nd toss is a 1, regardless of the 1st toss? 1/6*(5)*Are the above two events independent? No need to explain, just say yes or no. yes, the tosses don't affect each other.*(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)*What is the probability that the k-th toss is the first 1? (5/6)^(k-1) (1/6)

- 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 $},
*(5)*What is the probability of A? 1/4*(5)*What is the probability of B? 1/3*(5)*What is the probability of {$A\cap B$} ? 1/6*(5)*What is P[A|B]? 1/2*(5)*Are A and B independent? Why? no since 1/4 * 1/3 != 1/6

- There are 4 applicants for a 2 person team.
*(5)*How many different teams can I form? 4 choose 2 = 6*(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*(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*(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.

- 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.
*(5)*How many different ways can guides be assigned to tour groups? 4^3 = 64*(5)*One guide is named Able. What's the probability that Able is never picked (assuming randomness etc)? (3/4)^3 = 27/64*(5)*What's the probability that Able is picked exactly twice (assuming randomness etc)? (3 choose 2) * 3 / 64 =**9/64**

- 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.
*(5)*What percentage of the widgets that I buy are bad? .3*.2+.7*.1= .13 = 13%*(5)*Given I have a bad widget, what's the probability that it's from Acme? P(Acme|bad) = P(bad|Acme) P(Acme)/P(bad) = .2*.3/.13 = 6/13

*end of exam*

The first person finished in 26 min.