Probability Distribution/Random Variable problem.

[The_Iceflash]

Homework Statement

Roll a dice 4 times
X = smallest face observed
P(X=6) = 1/1296

Find:
P(X=1)
P(X=2)
P(X=3)
P(X=4)
P(X=5)

Homework Equations

N/A

The Attempt at a Solution

[PLAIN]http://img145.imageshack.us/img145/2773/62b19dc3faac4cc4a989485.gif

The issue is that the sum of numerators don't equal the denominator and I'm pretty sure it has to.

 

[jbunniii]

Your calculation for P(X = 1) is incorrect.

X = 1 iff at least one face is 1. The complement of this event is "no face is 1." So I can calculate the probability as follows:

P(X = 1) = 1 - P(\textrm{no face is 1}) = 1 - \left(5/6\right)^4 = 671/1296

Another, more laborious way to do this is to break the event down into disjoint cases that are easier to calculate:

"at least face is 1" is the disjoint union of the following:

"exactly one face is 1"
"exactly two faces are 1"
"exactly three faces are 1"
"exactly four faces are 1"

So you can calculate the probabilities of these four events and add them to get the answer.

The probability of exactly one face is

{4 \choose 1}(1/6)^1(5/6)^3

and so forth.

I didn't check your answers for P(X = 2) etc. You are correct that the probabilities of the 6 events you listed should add to 1.

 

[The_Iceflash]

^Thanks. What would I do to get P(X=2)...P(X=5)? I'm confused on what to do for the other faces.

 

[jbunniii]

For P(X = 2), I think you have to do disjoint cases as in my second method above.

"the smallest face is 2" is the disjoint union of the following events:

"exactly one face is 2, and the other faces are all > 2"
"exactly two faces are 2, and the other faces are all > 2"
"exactly three faces are 2, and the other faces are all > 2"
"all four faces are 2"

The probability of the first of these events is

{4 \choose 1}(1/6)^1(4/6)^3

and you can calculate the others similarly.

This method should work for P(X=3),...,P(X=5) as well. There might be a simpler short-cut, as there was with P(X = 1), but I don't see one.