How to find n in Binom. distr?

Question

In Restaurant Bijoux 13% of customers rated the food as ‘poor’, 22% of customers rated the food as ‘satisfactory’ and 65% rated it as ‘good’. A random sample of 12 customers who went for a meal at Restaurant Bijoux was taken. On a separate occasion, a random sample of n customers who went for a meal at the restaurant was taken. Find the smallest value of n for which the probability that at least 1 person will rate the food as ‘poor’ is greater than 0.95.

I got $$0.05>(0.87)^n$$ How do I solve it for n from here?

Answer 1

$0.05>0.87^n$ taking logarithms gives $\log (0.05)>\log(0.87^n)$ since $\log(x)$ is an increasing function then using rules of logarithms to bring the exponent down gives $\log (0.05)>n\log(0.87)$. Now note $\log(0.87)<0$ we divide by $\log(0.87)$ and so $21.511<n$ approximately. Hence since $n$ is the number of customers we must have $22<n$.

Answer 2

$$a = b^n$$

Hence

$$n = \log_b(a)$$

Which means

$$n = \log_{0.87}(0.05)$$

If you don't know how to calculate $\log_{00.87}$ you can always refer to the change of basis property, i.e.

$$\log_{b}(x) = \frac{\ln(x)}{\ln(b)}$$

Where $\ln$ means $\log_e$ where $e$ is the Euler Number (you can actually choose any $>1$ basis you like, but $e$ is more convenient).

Said this:

$$\log_{0.87}(0.05) = \frac{\ln(0.05)}{\ln(0.87)} \approx 21.51147352(...)$$

Eventually

$$n = 21.51143752(...)$$