Probability density function of dice

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SUMMARY

The discussion focuses on calculating the probability density function for the number of different faces shown when rolling a fair die three times. The probabilities for each outcome are established: for one face shown, the probability is 6*(1/6)^3; for two different faces, it is 6*5*3/216; and for three different faces, it is 6*5*4/216. The participants clarify the combinations and probabilities, ensuring that the total sums to one, confirming the calculations for each scenario.

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Homework Statement



Roll a fair die three times
Let X be the number of different faces shown all together ( X = 1,2,3 )
Find px(k)

Homework Equations





The Attempt at a Solution



Alright so I kno that i need to get the individual probabilities of each outcome
The first one where only one face is shown is (1/6)^3 * 6

The one where there are two different faces is (1/6)^2 * (5/6)

And I assume that likewise the final one with three different ones is (1*5*6)/(6^3)

Where I am getting stuck is I kno they have to sum to one so finding the different combinations to multiply to these expressions is giving me some trouble. For example the combinations in the same face was 6 because it could be either 1,2,3,4,5,6. Currently I have part 2 as having 30 different outcomes but it winds up looking a little off...any help would be great
 
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I'll agree with you one the first one. The first die can be anything and then the other two have to match. So (1/6)*(1/6). I don't agree on the third one. Again the first die can be anything but now the second can be any of the five remaining sides and the third can be any of the four remaining. For the second case, you should first account for the fact that any of the three dice can be the different one, and then multiply by the probability that the other two are different from that, but the same as each other.
 
yea thanks i figured it out

The first one is 6(1/6^3)

The second is 6*5 * 3 different combos over 216

and the last one is 6*5*4/ 216
 

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