Suppose Bob receives on average one call per night from his father.
Find the probability that 7th January was the third night this year (starting on 1st January) when no night calls were received.
I first calculated the probability of receiving no night calls for a particular night;
$$p_a = poissonpdf(1, 0) \approx 0.36788$$
Then, I used the Binomial probability density function along with some conditional probability to compute the required probability:
$$p = \frac{binompdf(7, 0.36788, 3)}{binompdf(6, 0.36788, 2)}$$
My idea behind this step is to find the probability of Bob receiving no calls for three days out of the 7 days, given that he has received no calls for two days out of the first six days. If he receives no calls for the two out of the first six days and for three out of the seven days, then definitely, he would have not received any calls for the third time on the seventh day.
However, the numerical solution disagrees significantly with my answer.
I'm very bad at probability. I think I'm getting some concepts wrong somewhere. Any help in understanding why this approach is wrong will be greatly appreciated. Thanks!
Let $X$ be the number of calls received in one night. $X$ is Poisson with parameter $\lambda=1$.
You are correct in the first part. The probability that $X=0$ is $\text{Pr}(X=0)=0.36788$.
I shall tackle the next part of your question. Your method needs some correction. What I think is that at the denominator, you did not consider the fact that a combination of any $2$ of the $6$ days could be also an event. You are only considering the combination but not the permutation.
What I would propose is to consider the Negative Binomial Distribution. Let event $NoCall$ be the event that there are no calls that day. Hence, $\text{Pr}(NoCall)=0.36788$. Let this also be considered a "success". As I am actuarial science trained, I will use the actuarial science notation for this, found here.
Let $Y$ be the number of getting $y$ failures prior to the $r$th success. In your case, every day from 1 Jan is a Bernoulli Trial where $Call$ is failure and $NoCall$ is success. So $Y$ is the number of $Call$ events before having $3$ $NoCall$ events.
I hope I have not confused you so far. Simply, the required probability is
$f_Y(y) = \binom{y+r-1}{r-1}(0.36788)^r(1-0.36788)^y$ where $y$ is the $y$th failure before the $3$rd success. In your case you want $y=4$. Simply plug in and you should get $0.11924$.