Binomial distribution of $P(x>8)$

Question

how do I do binomial distribution of $P(x>8)$, given $X\sim B (10,0.3)$

lets say for $P(X\ge2) = 1- p(X=0) - P(x=1)$

why is $P(x>8) = P(x=9) + P(x=10)$, how did this come about and why i cant do the same as above?

Answer 1

$\newcommand{\Prob}{\mathbb{P}}$You could indeed also do $\Prob(X > 8) = 1 - \Prob(X = 0) - \Prob(X=1)-\cdots - \Prob(X= 8)$, but it's just a lot more tedious since there are several more terms to calculate than if you just do $\Prob(X = 9) + \Prob(X = 10)$.

The reason that $\Prob(X > 8) = \Prob(X = 9) + \Prob(X = 10)$ is valid is that (assuming $X$ is Binomially distributed with $n = 10$), $X > 8$ occurs if and only if $X = 9$ or $X = 10$ (and these are two disjoint events so we can add their individual probabilities together).

Answer 2

If $X$ has binomial distribution with parameter $n=10$ then it takes values in $\{0,1,2,\dots,10\}$.

Now in the first place look at the events (i.e. the arguments of the probability measure $P$) that we are dealing with:

• $\{X\geq2\}=\{X<2\}^{\complement}=\Omega-\{X=0\}-\{X=1\}$
• $\{X>8\}=\{X=9\}\cup\{X=10\}$

Also we have:

• $\{X\geq2\}=\{X=2\}\cup\{X=3\}\cup\cdots\cup\{X=10\}$
• $\{X>8\}=\{X\leq8\}^{\complement}=\Omega-\{X=0\}-\{X=1\}-\cdots-\{X=8\}$

By calculation of the probability of these events of course we use the most convenient equality.