A car hire firm has three cars, which it hires out on a daily basis. Number of cars demanded per day follows a poisson distribution with mean 2.1

a) Find the probability that all cars are in use on any one day.

b) Find the probability that all cars are in use on exactly three days of a five day week.

What I tried:

For the a) part: normal poisson formula with mean = 2.1 and x = 3

$$\frac{(e-2.1)(2.1^3)}{6}$$

Got the answer as 0.1890. The book's answer is 0.350

For the b) part: did poisson to binomial. Found out p = 0.7. Then used 5C3. But I think my method here is wrong.

  • Let's start with a). Can you edit your question to show the "normal poisson formula" you used? (You can use MathJaX for formatting, if you know how to use it.) – Stephan Kolassa Sep 25 at 11:26
  • @StephanKolassa I don't know MathJax, so I avoided using formula in the question but I will try to write something out. – user585380 Sep 25 at 11:32
  • 1
    It says that it follows a Poisson distribution but your values are limited in [0,3], so it's not a real Poisson distribution, this is a Poisson binomial distribution. – user2974951 Sep 25 at 11:45
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    @user2974951, Number of cars demanded is Poisson, which is independent of number of cars you can serve. – gunes Sep 25 at 11:50
  • Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5 – user585380 Sep 25 at 12:11
up vote 4 down vote accepted

Let $X\sim Pois(\lambda = 2.1)$ be a Poisson random variable with mean $\lambda = 2.1$.

a) Let $A$ be the event all cars are hired in one day. Your mistake is only calculating the probability that $X=3$. In reality, if $X \geq 3$ then demand exceeds supply. So even though only $3$ cars are available, the probability is

$P(A) = P(X \geq 3) = 1 - P(X=0) - P(X=1) - P(X=2)=0.35$

b) If you can assume that the demand each day is independent, then using the binomial distribution is a correct method. There are $n=5$ days, each with probability $p=0.35$ of demand exceeding supply, so the probability that demand exceeds supply on exactly 3 days is given by

$5 \choose 3$$ 0.35^3 (1-0.35)^2$

ie, $P(Y=3)$ where $Y \sim Bin(5,0.35)$

  • 2
    I just did the same calculations after following some advice here and looked at the screen to see that you have done the same thing. Thank you so much for confirming. – user585380 Sep 25 at 11:58
  • No problem. Good luck with your studies :) – Xiaomi Sep 25 at 11:59
  • Thank you. If the question is, find the probability that exactly 10 cars are demanded in a five day week, binomial will be 15C10 (0.7)^10 (0.3)^5 – user585380 Sep 25 at 12:05
  • You can't use binomial in that situation, since it's no longer counting success/fails. For that question, use the fact that if cars demanded in 1 day is Poisson with rate 2.1, then cars demanded in 5 days is Poisson with rate $2.1\times 5 = 10.5$. – Xiaomi Sep 25 at 12:29
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    Binomial is when you have a fixed number of trials (eg. 5 days like in question b). But with "10 cars demanded in a 5 day week" the number of trials is not defined, because demands for cars are arriving at a random, not fixed, rate. Eg. You can't say "there was 100 times a car might have been demanded, but it was only demanded 10 times." The 5 days is not the number of trials in this instance because the success or failure is not determined on a daily basis (We are counting the total number of cars over all 5 days, not on each day) – Xiaomi Sep 25 at 13:49

Because, in (a), you need to calculate $P(X\geq 3)$, which is 0.35 nearly. If the number of requests coming is greater than 3, you still give your three cars. You don't have to have three requests exactly. For (b), you continue with classical binomial.

  • Is this supposed to be calculated from poisson to binomial. – user585380 Sep 25 at 11:56

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