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Expert: - 2/16/2009


Question
I'm having trouble grasping the "Probability Mass
Funtion"(pmf) and "Cumulative Distribution Function"(cdf)
concepts. Moreover I can't seem to figure out this problem.
I think I need an equation but I can't seem to put one
together with the given data. Here is the problem:

Two fair six-sided dice are tossed independently. Let M =
the maximum of the two tosses (so M(1,5) = 5, M(3,3) = 3,
etc.).

a)What is the pmf of M?[Hint:First determine p(1), then
p(2), and so on.]

b)Determine the cdf of M and graph it.(Don't need to do the
graph.)

Thanks for the help.

Answer
Two dice:
To get a 1, both dice must be a 1, so it's (1/6)(1/6) = 1 /36.
To get a 2 or less, it would be (2/6)(2/6), which would be 4/36.
...
To get a 5 or less, it would be (5/6)(5/6) = 25/36.
To get a 5 or less, it would be (6/6)(6/6) = 25/36.

So what we have for a cumulative distribution function
(the probability of getting no more than that number as a maximum)
is
N   P
1  1/36
2  4/36
3  9/36
4  16/36
5  25/36
6  36/36.

The probability distribution function would be the chance of getting
each number.  A 2 or less is 4/36.  A 1 is 1/36.  Therefore
4/36 - 1/36 = 3/36 of the time we actually get a 2 as the maximum.
Now if 1 has 1/36 chance and 2 has 3/35 chance, then the chance
of actually getting a 3 would be the chance of at least a 3 minus
(the chance of a 1 and the chance of a 2) = (9-3-1)/46 = 5/36.
In a similar fashion, we can get the chance of a 4 to be 7/36,
the chance of 5 to be 9/36, and the chance of a 6 to be 11/36.
The probability distribution is then
N  p
1  1/36
2  3/36
3  5/36
4  7/36
5  9/36
6  11/36.

Note that on the cumulative distribution, the last number is a 1
(36/36 is a 1).  On the probability distribution, the sum is 1.
That is, (1+3+5+7+9+11)/36 = 36/36 = 1.

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