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Let the random variable $W$ denote the number of students who take business calculus each semester at the University of Louisville. If the random variable $W$ has a Poisson distribution with parameter $\lambda$ equal to 300 and the probability of each student passing the course is $\frac{3}{5}$, then on an average how many students will pass the business calculus?

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$W$=Number of students taking the course. Say $X_{i}=1$ if the $i^{th}$ student passes the course and $0$ otherwise. So you have to find $E[\sum_{i=1}^{W} X_{i}]$.

Then $E[\sum_{i=1}^{W} X_{i}]=E[E[\sum_{i=1}^{W} X_{i}|W]]$. Now $E[\sum_{i=1}^{W} X_{i}|W=w]=\sum_{i=1}^{w}E[X_{i}]$ assuming independence of $X_{i}$ and $W$.

Now $E[X_{i}]=\frac{3}{5}$. Hence $E[\sum_{i=1}^{W} X_{i}|W=w]=\frac{3w}{5}\Rightarrow E[E[\sum_{i=1}^{W} X_{i}|W]]=E[\frac{3W}{5}]=\frac{3}{5}E[W]=\frac{3}{5}*300=180 \Rightarrow E[\sum_{i=1}^{W} X_{i}]=180.$

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Suppose that $X$ is the random variable denoting the number of students that pass the course each year. Then $X \sim B(\frac{3}{5}, W)$ where the right hand side denotes the binomial distribution with $W$ trials, and a probability of success of $\frac{3}{5}$ for each trial. Now since $$ \mathbb{E}(B(p, n)) = pn $$ we see that $$\mathbb{E}(X) = \frac{3}{5}\mathbb{E}(W) = \frac{3}{5}(300) = 180.$$ The last part is justified by the fact that for any Poisson distribution, so $\mathbb{E}(W) = \lambda = 300$, since the parameter of the Poisson distribution is defined to be the average events (in this case the number of people taking the course), which is what I used above.

Hopefully that helps.

I guess it does depend a lot on the rigour you require. The answer given by @AnirbanChatterjee is much better for this.

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Hint: Let $X\sim$ Poisson. What is $E[X]$?

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