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:

p₁ =  0

Step-by-step explanation:

$$\begin{gathered} p_0=1-0.1=0.9 \\ {P}_1=1-\left(\genfrac{}{}{0ex}{}{1}{0}\right)\cdot p_0^0\cdot (1-p_0{)}^{1-0}=1-(\genfrac{}{}{0ex}{}{1}{0})\cdot 0.9^0\cdot (1-0.9{)}^{1-0}=1\cdot 0.9^0\cdot 0.1^1=\frac{9}{10}=0.9 \\ {P}_2=1-\left(\genfrac{}{}{0ex}{}{2}{0}\right)\cdot p_0^0\cdot (1-p_0{)}^{2-0}=1-(\genfrac{}{}{0ex}{}{2}{0})\cdot 0.9^0\cdot (1-0.9{)}^{2-0}=1\cdot 0.9^0\cdot 0.1^2=\frac{99}{100}=0.99 \\ {P}_3=1-\left(\genfrac{}{}{0ex}{}{3}{0}\right)\cdot p_0^0\cdot (1-p_0{)}^{3-0}=1-(\genfrac{}{}{0ex}{}{3}{0})\cdot 0.9^0\cdot (1-0.9{)}^{3-0}=1\cdot 0.9^0\cdot 0.1^3=\frac{999}{1000}=0.999 \\ {p}_1=P_4=1-\left(\genfrac{}{}{0ex}{}{4}{0}\right)\cdot p_0^0\cdot (1-p_0{)}^{4-0}=1-(\genfrac{}{}{0ex}{}{4}{0})\cdot 0.9^0\cdot (1-0.9{)}^{4-0}=1\cdot 0.9^0\cdot 0.1^4=0.9999=0 \end{gathered}$$

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