Given $P(X=0)=P(X=3)=\frac18$ and $P(X=1)=P(X=2)=\frac38$, compute $P(X\geq 2)$ and $P(X>3)$

Question

Given $$X: \begin{pmatrix} 0 & 1 & 2 & 3 \\ \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \end{pmatrix}$$

Calculate $P(X \geq 2)$ and $P(X > 3)$

I'm learning for an exam and I create this example task on my own, I hope it's solvable at all. Can you please tell me if my solution is correct and if not, how to do it correctly?

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So for $P(X \geq 2)$ we can use the complementary probability

$$P(X \geq 2) = 1-P(X<2) = 1-\Big(P(X=0)+P(X=1)\Big)= 1-\frac{1}{8}-\frac{3}{8} = \frac{1}{2}$$

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For $P(X>3)$ we have

$$P(X>3) = P(X \geq 4) = 1-P(X<4) = \\ =1-\Big(P(X=0)+P(X=1)+P(X=2)+P(X=3)\Big) = \\ = 1-\frac{1}{8}-\frac{3}{8}-\frac{3}{8}-\frac{1}{8} = 0$$

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I doesn't necessarily need to make sense (maybe I created a bad task I mean) but it's very important for me to know if I solved it correct like that?

Answer 1

You are correct. However, you are making a little more work for yourself than needed.

Okay, the probability of the complement is sometimes easier to evaluate than the probability of the event itself.   But you have to judge when that is so.   This is a case when it is slightly less easy.

$\mathsf P(X\geq 2)~{=\mathsf P(X=2)+\mathsf P(X=3) \\= \tfrac 38+\tfrac 18\\=\tfrac 12}$

$\mathsf P(X> 3) =0$ simply because all values greater than three are not within the support for $X$.

But anyway, you arrived at the correct answers by a method that was not wrong.

Answer 2

Your solution is right, though maybe a bit laborious.

For $P(X\ge 2)$ you could just have said $P(X\ge 2)=P(X=2 \lor X=3)=P(X=2)+P(X=3)=\frac{3}{8}+\frac{1}{8}=\frac{1}{2}$. Note both you and I used the summation rule because the events $X=2$ and $X=3$ (or similar events $X=0$ and $X=1$) are mutually exclusive.

$X\gt 3$ cannot happen, so $P(X\gt 3)=0$.