A company expects its employees to finish a task at an average of 12 hours. What is the probability that 4 employees complete their tasks in a total of 10 hours or more?
Would this be a Poisson, binomial, or hypergeometric? It was posted by the instructor after learning about the aforementioned probability distributions
I am leaning toward poisson, solved in this manner:
$$ \lambda = 4 \text{ employees} * 12 \text{ average hours} = 48 \text{ hours} , $$ and the solution be $P(X > 10)$?
Thanks
It is important you understand that, in most practical situations, a Poisson random variable counts the number of times a particular event occurs in a specific interval of time or a specific region of space. These events must be independent, two events cannot occur simultaneously, and the average number of events that occur in a specific interval of time or a specific region of space is approximately constant. In your scenario, an "event" corresponds to an employee completing his or her work task.
That being said, if $X$ denotes the number of tasks a single employee finishes in a one hour period, then $X \sim \text{Poisson}(1/12)$ assuming aforementioned conditions are met.
If $T$ represents the time it takes for an employee to complete his or her first task, you can show that $T\sim \text{Exponential}(1/12)$. The argument goes like this. For $t> 0$ we have $$P(T\leq t)=1-P(T>t)=1-P\Big(\text{No tasks completed in } [0,t]\Big)$$
The number of tasks an employee finishes in $[0,t]$ is $\text{Poisson}(t/12)$ so $$P\Big(\text{No tasks completed in } [0,t]\Big)=e^{-t/12}$$ Hence $P(T \leq t)=1-e^{-t/12}$ which shows $T\sim \text{Exponential}(1/12)$.
Now assume that the employees complete their tasks independently of each other. Let $T_1,T_2,T_3,T_4$ denote the times it takes for the employees to finish their first tasks. You're looking to compute the probability that $\max\{T_1,T_2,T_3,T_4\}\geq 10$. We get $$P\Big(\max\{T_1,T_2,T_3,T_4\}\geq 10\Big)=1-P\Big(\max\{T_1,T_2,T_3,T_4\}< 10\Big)=1-\Big(1-e^{-5/6}\Big)^4$$