Homework Help: Finding the cumulative distribution function

1. Jul 21, 2010

cielo

1. The problem statement, all variables and given/known data

The experiment is to toss two balls into four boxes in such a way that each ball is equally likely to fall in any box. Let X denote the number of balls in the first box.

2. Relevant equations

What is the cumulative distribution function of X?

3. The attempt at a solution

I have enumerated what I suppose to be included in the sample space as follows: {(bb, 0, 0, 0), ((b, b, 0, 0), (b, 0, b, 0), (b, 0, 0, b), (0, bb, 0, 0), (0, b, b, 0), (0, b, 0, b), (0, 0, bb, 0), (0, 0, b, b), (0, 0, 0, bb)}

b stands for the box having a ball, bb stands for the box having the two balls, 0 for the box having no ball

I honestly don't know what to do next. Can some please guide me answering the problem?

2. Jul 21, 2010

praharmitra

Ok, your first step will be to draw up a probability distribution table. Now X denotes that the number of balls in first box. So, what is f(0), f(1), f(2), f(3), ... where f(x) is the probability that there are x balls in the first box. Since you have already listed all the possibilities of the distribution this shouldn't be too hard.

The cumulative distribution function of X, F(x) is defined as $$F(x) = \sum_{i = 0}^{i = x}f(i)$$

3. Jul 22, 2010

lanedance

though as there are 2 balls you only need consider up to X=2

4. Jul 23, 2010

cielo

Thank you for guiding me here.
I have constructed this table with the corresponding probability of values.
x 0 1 2
f(x) 6/10 3/10 1/10

Is there a general formula for finding the cumulative distribution function and the probability density function specific for this problem?

5. Jul 25, 2010

lanedance

those probabilties don't look right to me...

i think the best way for this case is a probability tree - consider separately throwing the 1st ball & the 2nd ball

In each case there is a 1/4 chance of a ball going into the 1st box, let "b" represent a ball in the first box, 0 not in 1st box

P(00) = (1/4)(1/4)
P(0b) = ...
P(b0) = ...
P(bb) = ...

see if you can fill in the other probabilties, in term of the random variable X
P(X=0) = P(00)
P(X=1) = P(0b) + P(b0)P(X=1)
P(X=2) = P(0b) + P(b0)

6. Jul 27, 2010

cielo

Praharmitra, thank you for the first one guiding me in this problem!

7. Jul 27, 2010

cielo

lanedance, you gave me a very idea in solving this problem.

P(x=0) = P(00)=(3/4)*(3/4)=9/16
p(X=1) = p(b0) + P(0b) = (1/4)*(3/4) + (3/4)*(1/4) = 6/16
P(x=2) = P(bb) = (1/4)*(1/4) = 1/16

now the cdf follows:
F(x) = o for 0>x
F(x) = 9/16 for 0<=x<1
F(x) = 5/16 for 1<=x<2
F(x) = 1 for x >=2

yehey

8. Jul 27, 2010

lanedance

nice work... but the cdf is cumulative

so the probs are correct in the pdf
p(X=0) = 9/16
p(X=1) = 6/16
p(X=2) = 1/16

the cdf should sum up, and i'd write it as follows
P(X<=0) = 9/16
P(X<=1) = 15/16
P(X<=2) = 1

9. Jul 27, 2010

cielo

Sorry, there were typo errors...excited in sharing what I've done. Once again, thank you! I sit here corrected.