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Calculations for Probability

1. For each of the following stats, prove the below or give counterexamples:
a. If 2 events A & B are independent, then A & B' are also independent.
b. Let's say that 3 events A,B,C have probabilities satisfying P(ABC)=P(A)P(B)P(C),then the events A & B are independent.
c. If P(A|B)>P(A), then P( A|B' )< P(A). (you may assume that 0<P(B)<1).
(PA|B) means Prob of A given B.)

2. 2 students in a class do not like waking up early, so they toss coins to decide who goes to class to take notes. Each student tosses a fair coin. If 1 student's coin shows heads while the others shows tails, then the student with heads wins (and doesn't go to class).If both student's coins shows the same thing, they toss again.
a. What is the probability that the students have to toss k times to determine the winner?
b. What is the probability that at least k tosses are needed to determine a winner?
c. Given that at least k tosses are needed to determine the winner, what is the probability that exactly k tosses are needed?
d. Is the event that k tosses are needed to determine a winner independent of the event that a particular player wins?
e. The students play this game for 3 straight lectures (Monday, Wed, Friday).Given that 21 tosses are needed altogether to decide the 3 games, what is the probability that exactly 7 tosses are needed for each of the 3 games?

3. A man likes to play tennis. When he hits the ball, the ball hits the court with probability 0.4 and is equally likely to hit any point on court. His partner hits the court with probability 0.9, & further it is twice as likely to hit the points on the back half of the court as the front half of the court.
a. Either the man or his partner hits the ball (but u don't know who) and the ball lands on the back right quadrant of the court. Is it possible to compute the probability that the man hit the ball, given this information? If not, what further info is needed?

4. In a kingdom, the royal family's eldest son becomes the king. Given that the king in this society has exactly 1 sibling, what is the probability that he has a sister?

5. When an experiment is run, a positive outcome occurs 30% of the trials.
Awe repeat this 20 times (each trial is independent).find the probability that the positive outcome occurs on k of the trials, for 0<=k<=20.
b. Again repeat 20 times. Given that at least 5 trials had positive outcome, what is the probability that exactly 7 trials had positive outcome?
c. Again repeat 20 times. Given that exactly 7 trials had positive outcome, what is the probability that the 17th trial had a positive outcome?
d. When we repeat this experiment 20 times (independently), what is the most likely number of successes?
e. We repeat this n times (independently).For what n is the probability of getting exactly 5 positive outcomes maximized.

6. We pick a person at random. The person selected will be a student with probability 0.8 and a faculty member with probability 0.2.
Faculty members have nothing to do on Friday evening with probability p (and are busy otherwise).Students have nothing to do with probability q (& are busy otherwise).A faculty member with nothing to do gets drunk with probability 0.8.A faculty member who is busy gets drunk with probability 0.1. A student with nothing to do gets drunk with probability 0.5. student who is busy gets drunk with probability 0.1.
a. What is the probability that a person chosen gets drunk on Friday night?
b. Given that a random person gets drunk, what is the probability that it is a faculty member?
c. Under what conditions on p and q is the event that a person gets drunk independent of the event that he is a faculty member?
d. Given that a randomly chosen person gets drunk, what is the probability that the person has nothing to do on Friday night?

Solution Preview

1. (a) True. Because P(A)=P(A(B+B'))=P(AB+AB')=P(AB)+P(AB')
=P(A)P(B)+P(AB'), then P(AB')=P(A)-P(A)P(B)=P(A)(1-P(B))
=P(A)P(B'). So A and B' are independent.
(b) True. P(ABC)=P(A)P(B)P(C) implies that A,B,C are mutually independent.
Thus A and B are independent.
(c) True. Because P(A)=P(A(B+B'))=P(AB+AB')=P(AB)+P(AB')
=P(A|B)P(B)+P(A|B')P(B')>P(A)P(B)+P(A|B')P(B')
then P(A|B')P(B')<P(A)-P(A)P(B)=P(A)P(B').
Thus P(A|B')<P(A)

2. We know, the probability that the two students' coins show the same thing is 1/2 (TT or HH); the probability that the two students' coins show different things is 1/2 (TH or HT) and in this case the game is over. Let n denote the number of times to determine the winner.
(a) P(n=k)=(1/2)^(k-1) * (1/2) = (1/2)^k
(b) P(n>=k)=(1/2)^k + (1/2)^(k+1) + ... =(1/2)^(k-1)
(c) P(n=k|n>=k) = ...

Solution Summary

This solution involves step by step calculations for probability.

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