1. Quiz 5 May 5 2005 5.1, 5.2 Discrete Probability

2. Quiz 4: May-04 ’04 3.30-3.45 pm In the following experiment we roll a fair die 5 times. a) What is the probability of the sequence “1,2,3,4,5”. b) What is the probability that the sequence starts with a “1”. c) What is the probability that the number “2” appears exactly twice. (hint: first count in how many ways you can place two “2”s in 5 slots, and then count the number of ways you can realize each such possibility). d) Let E be the event that we find the sequence “1,2,3,4,5” and let F be the event that the sequence starts with a “1”. What are the probabilities P(E|F) and P(F|E).

3. Quiz 4: Answers In the following experiment we roll a fair die 5 times. a) What is the probability of the sequence “1,2,3,4,5”. P = (1/6)^5 (each number has P=1/6 and all numbers are independent. Alternatively: there is 1 way to achieve this and 6^5 ways in S). b) What is the probability that the sequence starts with a “1”. P=6^4/6^5 = 1/6 (6^4 possible ways to achieve this and |S|=6^5). c) What is the probability that the number “2” appears exactly twice. (hint: first count in how many ways you can place two “2”s in 5 slots, and then count the number of ways you can realize each such possibility). P=C(5,2)x5^3/6^5 (C(5,2) is the number of ways to place two 2’s in 5 slots and each possibility can be achieved in 5^3 ways to fill the other slots with anything but not a 2.) d) Let E be the event that we find the sequence “1,2,3,4,5” and let F be the event that the sequence starts with a “1”. What are the probabilities P(E|F) and P(F|E). There is 1 solution in the intersection of E and F, while |F|=6^4. Hence P(E|F)=1/6^4. It is immediate that if E happened then F must be true: P(F|E)=1. Alternatively: Intersection E&F = 1, and |E|=1, therefore P(F|E)=1.