A boat rental company for tourism offers its clients two types of boats: sailboat and motorboat.
Furthermore, a client can take the PILOT option. In this case, the boat, whether sailboat or motorboat, is rented with a pilot.
We know that:
- $60\%$ of clients choose a sailboat; among them, $20\%$ take the PILOT option.
- $42\%$ of clients take the PILOT option.
A client is chosen at random and we consider the events:
- $V$: ``the client chooses a sailboat'';
- $L$: ``the client takes the PILOT option''.
Part A - Represent the situation with a probability tree that you will complete as you go.
- Calculate the probability that the client chooses a sailboat and does not take the PILOT option.
- Prove that the probability that the client chooses a motorboat and takes the PILOT option is equal to 0.30.
- Deduce $P_{\bar{V}}(L)$, the probability of $L$ given that $V$ is not realized.
- A client has taken the PILOT option. What is the probability that he chose a sailboat? Round to 0.01.
Part BWhen a client does not take the PILOT option, the probability that his boat suffers a breakdown is equal to 0.12. This probability is only 0.005 if the client takes the PILOT option. We consider a client. We denote by $A$ the event: ``his boat suffers a breakdown''.
- Determine $P(L \cap A)$ and $P(\bar{L} \cap A)$.
- The company rents 1000 boats. How many breakdowns can it expect?
Part CWe recall that the probability that a given client takes the PILOT option is equal to 0.42. We consider a random sample of 40 clients. We denote by $X$ the random variable counting the number of clients in the sample taking the PILOT option.
- We admit that the random variable $X$ follows a binomial distribution. Give its parameters without justification.
- Calculate the probability, rounded to $10^{-3}$, that at least 15 clients take the PILOT option.