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