From a book, I found these 2 questions which I have not understand.
(1) Suppose X is a discrete random variable with probability function
x 0 1 2 3 f(x) 27/64 27/64 9/64 1/64
Find the probabilities $P[\mu - 2\sigma < X < \mu + 2\sigma]$. Compare these probabilities with the probabilities that you will get from Chebyshev's Inequality.
When they solved the question, they used $\leq$ sign, that is $P[\mu - 2\sigma \leq X \leq \mu + 2\sigma]$. But the question is $P[\mu - 2\sigma < X < \mu + 2\sigma]$. Why they use the equal sign? Isn't in discrete case the equal sign's impact is tremendous?
Again when using Chebyshev's Inequality, they used $\leq$ sign, but in formula I know that it is $P[\mu - K\sigma < X < \mu + K\sigma]$.
(2) It is found from the Markov's inequality that for any non-negative random variable $X$ whose mean is $1$, the maximum possible value of $P[X \leq 100]$ is $0.01$.
But the formula of Markov's inequality is $$P[X \geq k] \leq E[X]/k.$$ Then how $P[X \leq 100]$ is $0.01$? In my consideration by using formula,
$$P[X \geq 100] \leq 1/100 \ \ \Rightarrow \ \ 1 - P[X < 100] \leq 1/100 \ \ \Rightarrow \ \ P[X < 100] \geq 1 - (1/100) = 0.99.$$
If this process is right how it conclude that $P[X≤100]$ is $0.01$?