I was wondering if anyone can please check this problem and let me know if I did this correct? Below is what i did to work out the problem. Thanks
Problem: The number of cars that arrive at a car wash follows Poisson distribution with the rate of 10.5 cars per hour. Assume that the number of cars that arrive during each hour are independent. (define a random variable for each part/problem)
d) Find the probability that there is at least one hour in a four hour period where only 5 cars arrive?
I believe this problem is both binomial and poisson distribution.
n= 4 , lambda= 10.5 , p=? Let Y represent the # of hours within a four hour period where only 5 cars arrive.
poisson distribution p= P(y=5)=poissonpdf(10.5,5)≈ 0.0293 therefore p=0.0293
binomial distribution P(x≥1) = 1 - P(x≤0)= 1- binomcdf(4,.0293,0)≈ 0.1121
Please let me know if I did any of these steps wrong! Any help would be appreciated. Thanks again
Yes. You are using that $(Y_k)\overset{iid}\sim\mathcal {Pois}(10.5)$ and $X\sim\mathcal{Bin}(4,\mathsf P(Y_1=5))$ to find $\mathsf P(X\geq 1)$.
Justification: $X$ is the count of successes in 4 independent trials, each with an identical success rate, where a trial's success is defined as obtaining exactly 5 arrivals from its Poisson process with arrival rate $10.5$ per the trial interval (one hour).
So as $\mathsf P(Y_1=5)=\dfrac{(10.5)^5 \mathsf e^{-10.5}}{5!}$, then $\mathsf P(X\geq 1)=1-\left(1-\dfrac{(10.5)^5\mathsf e^{-10.5}}{5!}\right)^4$.
That is indeed approximately $0.112101...$