A fair die is rolled until a number that has been obtained before is repeated. Let the random variable X denote the number of rolls. Compute the probability
mass function of X.
Here's how I tackled this problem. Could you please check if this true? If it is not could you please hint
We have that supp(X) is {2,3,4,5,6,7}
Then P(X=2)=(6/6)(1/6)= 1/6 because we have 1/6 choices that we obtain the previous number.
P(X=3)=(5/6)(2/6)=10/36. Because 5/6 is the second roll where it is the probability that we do not land the first value of the roll and 2/6 is the probability that we land the first two values of the rolls.
P(X=4)=(5.4.3)/6^3
P(X=5)=(5.4.3.4)/6^4
P(X=6)=(5.4.3.2.5)/6^5
P(X=7)=(5.4.3.2.1.6)/6^6
Best Answer
You can first evaluate $P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)+P(x=7)$ and it is $1$. So it is probably correct.
If $P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)+P(x=7)$ is not equal to $1$, then you must have calculated wrong.
This method can be used for checking.