I have two questions concerning the following: A pension (B&B) has three guest-rooms. The pension is opened from April - November (244 days, 35 weeks). Last year the distribution of the number of occupied rooms was as follows: 0 occupied rooms on 31 days, 1 occupied room on 22 days, 2 occupied rooms on 107 days and 3 occupied rooms on 84 days.
Now I'd have to calculate the probability that in a week (seven days) the pension is full for at least 5 days and state the assumptions you make.
I thought the following is correct, however I am not sure: The probability that the pension is full on a day is: $84/244 = 0.34$ Let $Y =$ number of days the B&B is full in a week. Then, $Y\sim Bin( n=7, p = 0.34)$ such that $P(Y > 4) = P(y=5) + P(y=6) + P(y=7) = \dots = 0.049$.
I am really not sure if this is correct?
Furthermore, suppose that the owner of the B&B has asked her daughter to come and visit her. This is only possible if there are no guests in the B&B. Calculate the exact 90% confidence interval for the number of days within she can visit her mother. (Use the exact probability distribution, so no Chebycheff)
$T$ = The Number of days within she can visit her mother.
Then $T \sim Geom(p)$ with $p = \text{no guests} / \text{total days} = 31/244 = 0.13$
So, I thought now you'd have to calculate the following interval: $P(c < T < d) = 0.90$.
I don't know how to go further without a known $c$ or $d$. So I would really appreciate your help!