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Here's the question:

"A sales representative vows to keep knocking on doors until he makes two sales. Given that his probability of success is $u$, let $X$ = the number of doors he knocks on.

Find the probability mass function of $X$"

My thought is that $x$ cannot be less than $2$, since he would have to knock on two doors to make two sales.

I'm thinking the function would be $\displaystyle\binom{x}{2} (u^2)(1-u)^{x-2}.$

But when I go to find $E(x)$, that doesn't lend itself well to the geometric form I've learned to love.

Am I on the right track at least?

Thanks for any help.

Khosrotash
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TrainChooseChoo
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3 Answers3

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Let us find the probability that $X=x$, that is, the probability she has to knock on $x$ doors.

She has to have $1$ success (and therefore $x-2$ failures) in the first $x-1$ trials, and then a success. The probability of this is $$\binom{x-1}{1}u(1-u)^{x-2}u.$$

André Nicolas
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  • I think I generalized correctly this logic to n successes $$\binom{x-1}{n-1}(1-u)^{x-n}u^n.$$ Interactive version here: https://www.desmos.com/calculator/ls9mbyrtul – Madacol Jan 14 '21 at 03:30
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The formula $$p_x=\binom x2 u^2(1-u)^{x-2}$$ is the probability that he knocks on at most $x$ doors. But you want the probability that he knocks on exactly $x$ doors. One way to find this would be:

  • number of possibilities for the first successful knock is $x-1$;
  • probability of each case is $u^2(1-u)^{x-2}$;
  • total probability is $(x-1)u^2(1-u)^{x-2}$.

Indeed it's not geometric, but instead of summing the series, you can use linearity of expectation: the expected value of $X$ will be the expected number of knocks to get one success, plus the number of additional knocks required to get the second success. And the latter is just the same as the former.

David
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The negative binomial distribution of trials until the second success is: $$\mathsf P(X_2=x) = {x-1\choose 1}u^2(1-u)^{x-2}$$

There must be $x-2$ failures and one success in any order before the last success.

The negative binomial distribution of trials until the r-th success is: $$\mathsf P(X_r=x) = {x-r+1\choose r-1}u^{r}(1-u)^{x-r}$$

The geometric distribution is the special case of a negative binomial where $r=1$.

$$\mathsf P(X_1=x) = {x\choose 0}u^{1}(1-u)^{x-1} = u(1-u)^{x-1}$$

Graham Kemp
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  • I believe your general case formula is incorrect, the binomial's upper parameter should be x-1 instead of x-r+1. Imagine 10 trials and 7 successes, the 7th success is at the 10th trial, but the previous 6 successes are in any combination inside the first 9 trials, so it should be 9 choose 6 instead of 4 choose 6 (which is impossible). – Madacol Jan 14 '21 at 03:54