Aryanna Blake

Answered

2022-10-21

Probability of two successes, possible geometric distribution problem
Let Y be the number of trials required to observe r successes. List the outcome space for the variable Y and show that the probabilities for Y are giver by the formula $P\left(Y=k\right)=\left(k-1\right)\left({p}^{2}\right)\left(1-p{\right)}^{k-2}$.
Okay I know the outcome space is $Y:=2,3,4,...$, the first two trials could be successes or you could never get two successes. The problem looks geometric but I dont undertand the extra $\left(k-1\right)\cdot p$ in front.

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Recalculate according to your conditions!

Answer & Explanation

Steinherrjm

Expert

2022-10-22Added 12 answers

Step 1
Assuming independence of trials and that each trial has probability p of success and also that $r=2$, as in the title:
$P\left[Y=k\right]$ is the probability that the second success occurred on the k'th trial. If the second success occurred on the k'th trial, then there was exactly one success in the first $k-1$ trials and the k'th trial was a success.
Now, there are $k-1$ ways to have exactly one success in the first $k-1$ trials (for example, one way is that the second trial was a success and the other $k-2$ trials were failures), each of which having probability $\left(1-p{\right)}^{k-2}p$. So the probability that exactly one success occurred in the first $k-1$ trials is
P(exactly one success in the first
Step 2
The probability of a success on the k'th trial is
The probability that the second success occurred on the k'th trial is the product of (1) and (2):
$P\left[Y=k\right]=\left(\phantom{\rule{thinmathspace}{0ex}}\left(k-1\right)\left(1-p{\right)}^{k-2}p\phantom{\rule{thinmathspace}{0ex}}\right)\cdot p=\left(k-1\right){p}^{2}\left(1-p{\right)}^{k-2}.$

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