Finding the PMF of a function of a discrete random variable

[Bill Headrick]

The discrete random variable K has the following PMF:

p(k) = { 1/6 if k=0
2/6 if k=1
3/6 if k=2
0 otherwise
}

Let Y = 1/(1+K), find the PMF of Y


My attempt:
So, I am really confused about what this is asking.

I took all of my possible K values {0, 1, 2} and plugged them into the formula for Y to get:

Y = {1,1/2,1/3}
Then the only Y value that is also a K value is 1 so:

p(y): {1 if y=1
0 ptherwise
}

This does not look right to me.

Am I approaching this the right way?

 

[mathman]

Your question is confusing (I don't know what PMF is supposed to be).

The density function for y:
p(y=1) = 1/6, p(y=1/2) = 1/3, p(y=1/3) = 1/2

 

[statdad]

PMF = probability mass function, since this is a discrete random variable the term "density" isn't typically used.
You don't want to base your probabilities for Y only on the ones for K. Think this way.
k = 0 if and only if Y = 1/(1+0) = 1, so p(y=1) = (fill in the blank)
K = 1 if and only if Y = 1/(1 + 1) = 1/2, so p(y = 1/2) = (again, fill in the blank)

Keep going this way.