1. ## Discrete probability

Firstly, I'm sorry if I'm posting an advanced question in the basic section, but I'm pretty sure this belongs here.

I took stats a few years ago and my course right now doesn't have a textbook so I'm not sure exactly how this is done.

Question reads: "The random variable X takes on one of the values 1,2,3,4,5 with probabilities
P(X=i) = ki, i = 1,2,3,4,5
for some value k. Find P(2 <= X <= 3)."

I'm not looking for a straight answer, just a question of where to start. I'm not sure but I thought that the sum of all P(X) across the domain of X (here [1,5]) should equal 1. So technically I should be able to solve for k right?

P(X) = kx
P(1) + P(2) + P(3) + P(4) + P(5) = 1.0
k + 2k + 3k + 4k + 5k = 1.0
14k = 1.0
k = 1/14
therefore:
P(X) = x/14

In which case the probability that X is 2 or 3 should be:

P(2 <= X <= 3) = P(2) + P(3) = 2/14 + 3/14 = 5/14

Is that right? or am I doing something entirely different?

2. Hello,

Perfect reasoning

except that 1+2+3+4+5=15, not 14

3. Oops... You are of course correct Moo. I'm really surprised my logic isn't in some way flawed (even if my implementation is).

Thanks a lot!

Question reads: "The random variable X takes on one of the values 1,2,3,4,5 with probabilities
P(X=i) = ki, i = 1,2,3,4,5
for some value k. Find P(2 <= X <= 3)."

I'm not looking for a straight answer, just a question of where to start. I'm not sure but I thought that the sum of all P(X) across the domain of X (here [1,5]) should equal 1. So technically I should be able to solve for k right?

P(X) = kx
P(1) + P(2) + P(3) + P(4) + P(5) = 1.0
k + 2k + 3k + 4k + 5k = 1.0
14k = 1.0
k = 1/14
therefore:
P(X) = x/14

In which case the probability that X is 2 or 3 should be:

P(2 <= X <= 3) = P(2) + P(3) = 2/14 + 3/14 = 5/14
You have a small error. $\displaystyle k=\frac{1}{\color{blue}15}$

5. Thanks for the reply Plato, but I think Moo beat you to the punch