How to calculate probability and expected value?

The distribution function of a discrete random variable X is given

$F_X(x)=\begin{cases} 0, &x<1\\ \frac{5}{13},& 1\leq x< 2 \\ \frac{10}{13}, & 2\leq x<3 \\ \frac{11}{13}, & 3\leq x<4 \\ 1, & 4\leq x \end{cases}$

$A=(X=2)\cup (X=4)$

Calculate: $P(A)$ and $E(X)$

I was thinking to solve $P(A)$ with formula: $P(a)=\begin{pmatrix} n \\ a \end{pmatrix} p^a (1-p)^{n-a}$, but I dont $p$ and $n$. Which formula I should use?

• Can you not try to apply random formulas and just find the values this variable takes and corresponding probabilities (and then use some general formula)? – Sergei Golovan Dec 25 '16 at 18:35
• Is $X$ a discrete or a continuous random variable? To begin, can you find $P(X=2)?$ – BruceET Dec 25 '16 at 18:36

the random variable $X$ can take four values, which are exactly the points of discontinuity of $F_X$: $$\mathbb P (X=1)= \frac 5 {13}, \quad \mathbb P (X=2)=\frac {10} {13}- \frac 5 {13}, \quad \mathbb P (X=3)=\frac {11} {13} -\frac {10} {13}, \quad \mathbb P (X=4)=1- \frac {11} {13}. \quad$$ Therefore $$\mathbb P(A)=\mathbb P(X=2)+\mathbb P(X=4) = \frac 7 {13},$$ and $$\mathbb E [X]= 1 \cdot \frac {5}{13}+2 \cdot \frac {5}{13}+3 \cdot \frac {1}{13}+4 \cdot \frac {2}{13}=2.$$

Hint:

$F_X(x) = P(X \leq x)$

$(X =2 ) = (X\geq2) \cap (X>2)^c$

For a discrete random variable $X$,

$$E[X] = \sum_{i=1}^n x_i P(X=x_i)$$

Note that $P[X=t] = P[X \le t] - \lim_{x \uparrow t} P[X \le x]= F(t) - \lim_{x \uparrow t} F(x)$.

The limits are particularly straightforward to compute since $F$ is piecewise constant.