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 $Y \sim Bin(5,0.35)$
You would pass in the sum of your observations, and set the T parameter in poisson.test to the number of days you have taken samples:
#pretend these are your data points
x <- 8:16
poisson.test(sum(x),T=length(x),r=10)
# > Exact Poisson test
# >
# > data: sum(x) time base: length(x)
# > number of events = 108, time base = 9, p-value = 0.0647
# > alternative hypothesis: true event rate is not equal to 10
# > 95 percent confidence interval:
# > 9.843833 14.488079
# > sample estimates:
# > event rate
# > 12
Upping the T parameter reduces the variance of the overall estimate. See for example if we take the mean rate instead (and pass T=1):
poisson.test(mean(x),T=1,r=10)
# > Exact Poisson test
# >
# > data: mean(x) time base: 1
# > number of events = 12, time base = 1, p-value = 0.5234
# > alternative hypothesis: true event rate is not equal to 10
# > 95 percent confidence interval:
# > 6.200575 20.961585
# > sample estimates:
# > event rate
# > 12
So repeating the counts you gain precision in your estimate.
Since you can see that this only works with the overall sum of the observations, you will also likely want to check and see that the individual observations approximately look like a Poisson distribution as well (e.g. if everyday is exactly 10, then it is not Poisson).
I have a package I am working on, ptools, where you can see the Poisson fit to data using the check_pois function.
Best Answer
The answer to your first question is yes. If the sum of two independent variables is poisson then the individual variables are also poisson. See Raikov's Theorem