Conditional probability and independence using Bayes' rule and binomial expansion

Question

I would like to start the question by adding some context.

Tom is a night owl and is very hungry at night. On day , he eats at home with probability 1− (0<<1); or, with probability , he goes out and goes the -th restaurant in his city. At any restaurant that he goes to, either:

(i) the restaurant is open with probability (where 0<<1). In that case, Tom orders some food for take-away;

(ii) the restaurant is closed so Tom returns home and has sleep for dinner because he is sad.

We assume that the collection of all events of the form {Tom stays home on day } and {The -th restaurant is open on day }, for =1,2,…, are (mutually) independent.

The probability that Tom orders some food for take-away on the first day is $p*q$. So simple?

Furthermore, fixing some integers and , with ≥1 and 0≤≤.

The probability that he ordered food for take-away exactly times on the first days is

$${n \choose k}q^k(1-q)^{n-k}=\frac{n!}{k!(n-k)!}*q^k(1-q)^{n-k}$$

This doesn't seem right to me. What about $p$? Tom only orders food if he goes out, which is probability of $p$.

The conditional probability of Tom staying home on day 1, given that he did not order food on that day is

$$\frac{1-p}{p*(1-q)}$$

Where $p*(1-q)$ is basically the he went out and the restaurant was closed. But he can also decide to stay home with probability of $1-p$ which means he doesn't order food.

My actual question is,

Within the first week (7 days), what is the probability that there were no days in which Tom went out but did not order food from a restaurant?

That conjunctive "but" in the question confuses me. If he doesn't go out in 7 days in a row, I would say $(1-p)^7$

I don't know, this all feels wrong. Please help me and feel free to correct my reasoning.

---

EDIT

Tom ordering food for take-away exactly times on the first days should be

$${n \choose k}(pq)^k(1-pq)^{n-k}=\frac{n!}{k!(n-k)!}*(pq)^k(1-pq)^{n-k}$$

Answer 1

The correct interpretation is

• Tom goes out
• Tom doesn't order food

Mathematically, that's all we need to know. Hence

$${7 \choose 0}((1−))^0(1-(1−))^{7-0}=(1-(1−))^{7}=(1−+)^{7}$$

Where $(1−)$ denotes the probability for the interpretation above.

Answer 2

To address your "actual question" , I interpret the wording to mean that on each day that he goes out, the restaurant is open so he can and does order food. [= On no day that he goes out does he not order food]

P(Tom goes out on $k$ days in a week) = $\binom7{k}p^k(1-p)^{7-k}$

P(restaurant open on each of these particular $k$ days) $= q^k$

Thus indicated $Pr = \binom7{k}(pq)^k(1-p)^{7-k}$