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In an urn there are $u$ balls, $b$ of which are black.

If we perform $n$ trials of one ball at a time with replacement, the probability of the event $E$ to get $n$ times a black ball is $P(E)=\left(\frac{b}{u}\right)^n$, whereas the probability of the event $L$ to get at least one black ball is $P(L)=1-\left(\frac{u-b}{u}\right)^n$.

Let $X$ be the non-negative, integer-valued random variable representing the number of trials it takes to get a success for the event $E$, and $Y$ the non-negative, integer-valued random variable representing the number of trials it takes to get a success for the event $L$.

What are the expected values $\mathbb{E}[X]$ and $\mathbb{E}[Y]$?

My attempt for $\mathbb{E}[X]$:

I denote with $P(E;k)$ the probability to get a success for the event $E$ at the trial $k$.

The variable $X$ can assume only the value $n$, therefore, by definition it should be $$\mathbb{E}[X]=\sum_{k=1}^n k P(E;k)=\sum_{k=1}^n k P(E)\delta_{n,k}=nP(E).$$

My attempt for $\mathbb{E}[Y]$:

I denote with $P(L;k)$ the probability to get a success for the event $L$ at the trial $k$.

The variable $Y$ can assume any value from $1$ to $n$, therefore, by definition, it should be $$\mathbb{E}[Y]=\sum_{k=1}^n k P(L;k)=\sum_{k=1}^n k \left[1-\left(\frac{u-b}{u}\right)^k\right].$$

Are these calculations correct?

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    $\begingroup$ If I understand your question (the desription does not make much sense for me), the random variables $X$ and $Y$ are geometrically distributed. These random variables are not bounded by $n$, so your calculations cannot be correct. Please google for geometric distribution. $\endgroup$ – Dieter Kadelka Jan 29 at 18:26
  • $\begingroup$ @DieterKadelka Thanks for your comment. $\endgroup$ – Andrea Prunotto Jan 29 at 18:52
  • $\begingroup$ In your definitions of of $X$ and $Y$, what do you mean by "trials" and "a success for the event"? $\endgroup$ – Iosif Pinelis Jan 30 at 2:42
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    $\begingroup$ Sorry, I still don't understand your definitions. Perhaps you can state them in purely formal terms (such as independent identically distributed random variables, with certain distributions) and completely eschew such non-mathematical terms as "draw", "attempt", "trials", etc. $\endgroup$ – Iosif Pinelis Jan 30 at 17:32
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    $\begingroup$ I'm afraid the question as phrased has no answer; a different question does: "Repeatedly draw a ball with replacement, stop when you have $n$ black balls. What is the expected number of balls drawn?" --- is that a question that would interest you? $\endgroup$ – Carlo Beenakker Feb 5 at 9:10
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For the alternative approach in the comments, the Pascal distribution applies: Repeatedly draw a ball with replacement from an urn with $u$ balls, $b$ of which are black, stop when you have $n$ black balls. Let $X$ be the number of balls drawn. The probability distribution of $X$ is $$P(X=N)=\binom{N-1}{n-1} (b/u)^{n}(1-b/u)^{N-n},\;\;N\geq n,$$ with expectation value $\mathbb{E}(X)=nu/b$.

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  • $\begingroup$ Thanks Carlo, it looks really neat! Please, can you help me to apply the same approach to the other event? $\endgroup$ – Andrea Prunotto Feb 5 at 18:15
  • $\begingroup$ For event $Y$ the best I can come up with is to ask for the number of draws until the first black ball, which has expectation value $u/b$. $\endgroup$ – Carlo Beenakker Feb 5 at 22:01
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In your definitions, you have confused:

$$P(E;k)$$

This actually means:

$$P(X=k|E)$$

(which equals $\delta_{kn}$ and gives the correct value that $\mathbb E(X)=n$).

and not, as you have used:

$$P(E|X=k)$$

which equals $P(E)\delta_{kn}$.

See Conditional Probability for more information.

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