Poisson and conditional probability

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?

• 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. – jbowman Dec 21 '18 at 16:00
• Sorry, it was a typo. Fixed. – David Duarte Dec 21 '18 at 16:31
• 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? – whuber Dec 21 '18 at 16:53
• 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. – Michael Hardy Dec 21 '18 at 22:42
• $\ldots\,$also, notice that those events are NOT independent, so you can't just multiply their probabilities to get the probability of their intersection. – Michael Hardy Dec 21 '18 at 22:44

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