Probability - R successes before the kth failure

In summary, the probability of success is p and the probability of having at least r successes before the k-th failure is p(k+r+1).
  • #1
dkotschessaa
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Homework Statement



In a sequence of independent trials with probability of success p, what is the probability that there are r successes before the k-th failure?

Homework Equations



Binomial distribution

[itex] f(x;n,p) = {{n}\choose{r}} p^{x} (1-p)^{n-r} [/itex]

The Attempt at a Solution



I know that the answer is

[itex]{{k+r+1}\choose{r}} p^{r} (1-p)^{k} [/itex]

This almost makes sense to me.

Since x = number of successes = r
and I know n represents the number of trials - but I'm not sure how this amounts to k+r+1.

and if n is k+r+1, then the exponent of (1-p), which is n-x, should be

(k+r+1) - r = k+1

But here it is just k.

why?

-Dave K
 
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  • #2
dkotschessaa said:

Homework Statement



In a sequence of independent trials with probability of success p, what is the probability that there are r successes before the k-th failure?

Homework Equations



Binomial distribution

[itex] f(x;n,p) = {{n}\choose{r}} p^{x} (1-p)^{n-r} [/itex]

The Attempt at a Solution



I know that the answer is

[itex]{{k+r+1}\choose{r}} p^{r} (1-p)^{k} [/itex]

This almost makes sense to me.

Since x = number of successes = r
and I know n represents the number of trials - but I'm not sure how this amounts to k+r+1.

and if n is k+r+1, then the exponent of (1-p), which is n-x, should be

(k+r+1) - r = k+1

But here it is just k.

why?

-Dave K

The "answer" you give above is not quite correct. Look at it for the case of k = 1; it does not work properly in that case. (Note: your formula would be the probability of having r successes and 1 failure in (r+1) trials, but that would include all outcomes of the form FSS...S, SFSS...S, ... SS...SF, which is NOT what you want. You want the probability of the single point SSS...SF.)

The number of tosses X until the first failure is geometrically distributed with parameter 1-p; note that this count *includes* the failure itself, so the number *before* the first failure is Y = X-1 (which is in {0, 1, 2, ...}). After each failure the counting process starts again, so you want the distribution Zk = Y1 + Y2 + ... + Yk, where the Yi are iid copies of Y. You are asking for P{Zk = r}. You can get a closed-form formula for this using generating-function methods, or else by getting a recursion for P{Z(k-1)=s} and P{Y=t}. The answer will be almost what you wrote above, but with a slight difference.
 
Last edited:
  • #3
Thanks. I suspected the answer was wrong, but it was what the professor gave us. He is giving us wrong answers, and sometimes even wrong questions. We are all baffled. We have a test tomorrow.

Thanks!
 

1. What is "Probability - R successes before the kth failure"?

"Probability - R successes before the kth failure" is a mathematical concept used to calculate the likelihood of achieving a certain number of successes before a specific number of failures in a series of trials. It is often used in situations where there is a desired outcome and a set number of attempts to achieve it.

2. How is the probability calculated for "R successes before the kth failure"?

The probability for "R successes before the kth failure" is calculated using the binomial distribution formula, which takes into account the number of trials, the probability of success in each trial, and the desired number of successes and failures. It can also be calculated using a probability tree or by using a calculator or software program.

3. What factors can affect the probability of "R successes before the kth failure"?

The probability of "R successes before the kth failure" can be affected by several factors, such as the number of trials, the probability of success in each trial, and the desired number of successes and failures. Other factors that can influence the probability include the randomness of the events, the independence of each trial, and any external factors that may impact the outcome.

4. Can the probability of "R successes before the kth failure" be greater than 1?

No, the probability of "R successes before the kth failure" cannot be greater than 1. This is because a probability of 1 represents a 100% chance of the desired outcome occurring, and a probability greater than 1 would imply a certainty of success before the kth failure, which is not possible in most situations.

5. How can "R successes before the kth failure" be applied in real-life situations?

"R successes before the kth failure" can be applied in various real-life situations, such as predicting the success rate of a marketing campaign, estimating the number of defective items in a production line, or determining the likelihood of getting a certain number of correct answers on a multiple-choice test. It can also be used in gambling and sports betting to calculate the probability of winning a certain number of bets before reaching a certain number of losses.

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