Admit that the number of participants who intend to enroll in a given training follows a Poisson distribution with a mean of $12.$ If there is not a minimum of five enrollments, training is not offered. On the other hand, the room where the training takes place has a maximum of $20$ participants.

Calculate the probability that all the interested parties will be able to register, since the minimum number of registrations has been reached.


As we have already achieved 5 entries we can only accept 15 more. So, the answer would be:

$$ P(k = 15) = \frac{e^{-12} 12^{15}}{15!} $$

In the other hand, using conditional probability formula:

$$ P\left( k \le 20\mid k \ge 5 \right) = \frac{{P(k \le 20) \cap P(k \ge 5)}}{P(k \ge 5)} = \frac{{P(k \le 20) \cdot P(k \ge 5)}}{P(k \ge 5)} $$

$$ P(k \le 20\mid k \ge 5) = P(k\le20) = \sum_{k=0}^{20} \frac{e^{-12} 12^k}{k!} $$

My question is, which of approaches are correct? Why?

  • $\begingroup$ Note that in your first formula the number "15" never appears. This implies that $P(k=15) = P(k=1) = P(k=105)\cdots$, which is obviously wrong. $\endgroup$ – jbowman Dec 21 '18 at 16:00
  • $\begingroup$ Sorry, it was a typo. Fixed. $\endgroup$ – David Duarte Dec 21 '18 at 16:31
  • $\begingroup$ I am unable to see how the event "$k=15$" (which as we may infer from your formula for it in the first answer, means there are exactly $15$ interested parties) is directly relevant to answering the question, given that all interested parties will be able to register even when there are fewer than $15$ interested parties or between $16$ and $20$ interested parties. Can you explain? $\endgroup$ – whuber Dec 21 '18 at 16:53
  • 1
    $\begingroup$ It is nonsense to write $$ P(k\ge5)\cap P(k\le20) =\cdots. $$ That should instead say $$ P(k\ge 5 \cap k\le 20) = \cdots. $$ You can take intersections of events; you cannot take intersections of numbers. $\endgroup$ – Michael Hardy Dec 21 '18 at 22:42
  • $\begingroup$ $\ldots\,$also, notice that those events are NOT independent, so you can't just multiply their probabilities to get the probability of their intersection. $\endgroup$ – Michael Hardy Dec 21 '18 at 22:44

I don't think the poisson is memoryless (for if it were, the two solutions would be equivalent, and the question would be moot). Note the problem says that the minimum number of registrations has been reached. It doesn't say exactly 5 have been reached, so we have to assume at least 5 have been made.

I would recommend computing

$$P(k\leq 20 \vert 5\leq k) = \dfrac{P(5\leq k \leq 20)}{1-P(5>k)} $$


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