EXpectation value of the Poisson distribution

Question

I've this homework problem which I do not understand how to solve. A book has 153 pages and number of typos on each page is a Poisson random variable with mean $\lambda$, and is independent of the number of typos on all other pages. What is the expected number of pages that do not have a mistake on them?
The expected value of Poisson distribution is
$$
P(k;\lambda) = \frac{e^{-\lambda} \lambda^{k}} {k!} \quad \langle k \rangle = \lambda = NP \;\; \text{similar to bionomial distribution}
$$
Can someone please provide some hints?

Answer 1

Let $X_i$ be the random variable describing how many typos there are on page $i$. Then $X_i \sim Poisson(\lambda)$. Then define the random variable $Y_i$ by $$ Y_i := \begin{cases} 1,\ &\text{if}\ X_i = 0, \\ 0,\ &\text{if}\ X_i>0.\end{cases} $$ The random variable $Y_i$ describes whether or not there is a typo on page $i$.

Then $Y := \sum_{i=1}^{153} Y_i$ is a random variable describing the number of pages that do not have typos on them. How can you compute $\mathbb{E}(Y)$? The expectation function $\mathbb{E}$ is linear, so: $$ \begin{align} \mathbb{E}(Y) &= \mathbb{E}\bigg(\sum_{i=1}^{153} Y_i\bigg) \\[1ex] & = \mathbb{E}(Y_1+Y_2+Y_3+\ldots+Y_{153})\\[1ex] & = \sum_{i=1}^{153} \mathbb{E}(Y_i) \\[1ex] & = \mathbb{E}(Y_1) + \mathbb{E}(Y_2) + \mathbb{E}(Y_3) + \ldots + \mathbb{E}(Y_{153}) \end{align}$$ For each $Y_i$, we know $\mathbb{E}(Y_i)=\mathbb{P}(X_i=0)$. Put this together to find your answer.