0

The following question is from Blitztein-Hwang, Introduction to Probability:

Independent Bernoulli trials are performed, with probability 1/2 of success, until there has been at least one success. Find the PMF of the number of trials performed.

Let $X$ be the number of trials performed.

When I first attempted the question, I concluded that $P(X=k) = 1-(\frac{1}{2})^k$ via the principle of inclusion-exclusion and the independence of events. My thought process here was the following: because the question says until there has been at least one success, then for $k$ trials, we need at least one success, i.e., we can get up to $k$ success.

Then, after a quick search, I stumbled on this previous question: Bernoulli trials with at least 1 success and 1 failure. In the answer, it is stated that $X$ has a geometric distribution, i.e., $P(X=k) = (1-\frac{1}{2})^{k-1}\cdot \frac{1}{2} = (\frac{1}{2})^k$, but this assumes that after $k$ trials, we only have one success, so it seems more close to until there has been a success and not until there has been at least one success

Please, can someone help me understand which case better reflects the question and why?

vxek
  • 197
  • This is not clear. What is $X$? The first success? But that doesn't make sense...you have $P(X=k)\to 1 $ for large $k$. If nothing else, we would have to have $\sum_k P(X=k)=1$. Your sum diverges. – lulu Aug 31 '22 at 21:04
  • $X$ is the random variable that keeps track of the number of trials performed – vxek Aug 31 '22 at 21:24
  • That's not a variable. You mean the number of trials it takes until you observe the first success? Something else? – lulu Aug 31 '22 at 21:27
  • Based on what the problem said, I will say that it is the number of trials it takes until I observe at least one success... – vxek Aug 31 '22 at 21:56

1 Answers1

2

Your error is in this line:

My thought process here was the following: because the question says until there has been at least one success, then for k trials, we need at least one success, i.e., we can get up to k success.

The key detail is that you're counting until the first success. This doesn't mean you can have up to $k$ successes; it means instead that your first $k-1$ trials could not have had a success, but that your $k^{\text{th}}$ one did, making it the first of them. That's the role of the word until in this context; it implies that we stop when we find a success. (Yes, the use of "at least one" becomes necessarily strange in this formulation.)

Aaron Montgomery
  • 10,688