Bernoulli trial

A used car saleswoman estimates that each time she shows a customer a car, there is a probability of 0.1 that the customer will buy the car. The saleswoman would like to sell at least one car per week. If showing a car is a Bernoulli trial with a probability of 0.95 of at least one sale,
(i) how many cars would the saleswoman show per week?
(ii) calculate the probability of exactly three sales per week.

Correct answer:

n =  29
p₃ =  0.2361

Step-by-step explanation:

$$\begin{gathered} k=0.1 \\ {p}_0=1-k=1-0.1=\frac{9}{10}=0.9 \\ 1-p_0^n=0.95 \\ {p}_0^n=0.05 \\ n\ln p_0=\ln 0.05 \\ {n}_0=\frac{\ln 0.05}{\ln (p_0)}=\frac{\ln 0.05}{\ln 0.9}\doteq 28.4332 \\ n=\lceil n_0\rceil =\lceil 28.4332\rceil =29 \end{gathered}$$

$p_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot k^3\cdot (1-k{)}^{n-3}=(\genfrac{}{}{0ex}{}{29}{3})\cdot 0.1^3\cdot (1-0.1{)}^{29-3}=3654\cdot 0.1^3\cdot 0.9^{26}=0.2361$

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