Help with negative binomial distributions

AI Thread Summary
To find P(X ≥ 3) for a negative binomial random variable with p = 0.6, it's essential to understand that the distribution models the number of trials until the r-th success. For r = 2, the correct approach is to calculate 1 - P(X ≤ 2), noting that P(0) and P(1) are zero, leading to P(X ≥ 3) not being zero. For r = 4, P(X ≥ 3) equals 1, as achieving four successes requires at least four trials. Clarification from the teacher is recommended due to discrepancies in the provided answers. Understanding the properties of the negative binomial distribution is crucial for accurate calculations.
mintsharpie
Messages
3
Reaction score
0
One of the questions in my probability homework reads:

X denotes a negative binomial random variable, with p = 0.6 Find P(X ≥ 3) for a) r = 2 and b) r = 4.

According to my teacher, the answers are 0.1792 and 0.45568, respectively, but I can't for the life of me figure out how he got them. I tried finding P(X ≥ 3) by turning it into 1 - P(X ≤ 2) and then calculating p(2), p(1), and p(0), but I kept getting 0 for my answer, which obviously isn't correct.

Can someone please help me solve this problem, or explain to me how I would go about solving it? I'm really confused.

Thanks.
 
Physics news on Phys.org
mintsharpie said:
One of the questions in my probability homework reads:

X denotes a negative binomial random variable, with p = 0.6 Find P(X ≥ 3) for a) r = 2 and b) r = 4.

According to my teacher, the answers are 0.1792 and 0.45568, respectively, but I can't for the life of me figure out how he got them. I tried finding P(X ≥ 3) by turning it into 1 - P(X ≤ 2) and then calculating p(2), p(1), and p(0), but I kept getting 0 for my answer, which obviously isn't correct.

Can someone please help me solve this problem, or explain to me how I would go about solving it? I'm really confused.

Thanks.

Remember that the negative binomial models the number of Bernoulli trials up to and including the rth success. Therefore p(x) > 0 only for x \ge r.

If r = 2, then p(0) and p(1) are obviously zero, so your first calculation should just be 1 - p(2), which isn't zero and isn't his answer either. And if r = 4, obviously P(X ≥ 3) = 1 since you can't have 4 successes in less than three trials. Time to ask your teacher what's going on.
 
Back
Top