Let X be a Poisson random variable with mean 20. Let $p=Pr[X≥26]$.
a) Use the Markov inequality to obtain an upper bound on p.
b) Use the one-sided Chebyshev inequality to obtain an upper bound on p.
c) Use the Chernoff bound to obtain an upper bound on p.
d) Approximate p using the central limit theorem.
For part a I took the Markov inequality to calculate $Pr[X\ge a]=\frac{E[x]}{a}=\frac{20}{26}$, and simplifying the fraction is trivial. For part b I applied $Pr[|X-a|\ge k] \le \frac{\sigma^2}{k^2}$, assuming $k=26$, which leads to $\frac{100}{169}$. Are these solutions correct? I'm not quite sure how to take care of parts c and d.
Since Poisson distributions are self-similar, the random variable $X$ can be expressed as a sum of an arbitrary number ($n$) of independent Poisson distributed random variables. $$X = X_{1} + X_{2} + \cdots + X_{n} \ ,$$ where the expected value $$\mathbb{E}\{X_{k}\} = \frac{20}{n} \ , \quad k \in \{1,2,\ldots,n\} \ .$$ Using this representation of $X$ it is straightforward to apply the Central limit theorem to approximate the probability $\mathbb{P}(X\geq 26)$.
As I understand it, in order to apply a Chernoff bound on the probability $\mathbb{P}(X\geq 26)$ the random variable $X$ needs to be expressed as a sum of binary random variables; the random variables $X_{k}$ are not binary. However, you can make use of the connection between Poisson distributions and Binomial distributions, and the fact that Binomial distributions can be viewed as sums of binary random variables.
Thus if $Y$ is Binomial distributed $\text{Bin}(m,q)$ where $m$ is large and $q$ is small (and their product $mq = 20$) then $Y$ can be expressed as $$Y = Y_{1}+Y_{2}+\cdots+Y_{m}$$ where $Y_{k}$ is binary (0 or 1) and $$\mathbb{P}(Y_{k} = 1) = q\ , \quad k\in\{1,2,\ldots,m\}\ .$$ Using the approximation $\text{Bin}(m,q) \approx \text{Poi}(20)$ the probability $$\mathbb{P}(X\geq 26) \approx \mathbb{P}(Y\geq 26)\ .$$ The probability $\mathbb{P}(Y\geq 26)$ can be bounded using a Chernoff bound.