<s>[INST] Expected value and probability 
The distribution function of a discrete random variable X is given
$$F_X(x)=\begin{cases} 0, &x<3\\ \frac{6}{11},& 3\leq x< 4  \\ \frac{7}{11}, & 4\leq x<5 \\ \frac{8}{11}, & 5\leq x<6 \\ 1, & 6\leq x   \end{cases} $$
Let $A=(X=3)\cup (X=5)$. Calculate: $P(A)$ and $E(X)$

How I did it:
$P(1)=0, P(2)=\frac{6}{11}, P(3)=\frac{1}{11}, P(4)=\frac{1}{11}, P(5)=\frac{3}{11}$
$P(A)=\frac{1}{11}+\frac{3}{11}=\frac{4}{11}$ (Correct solution: $\frac{7}{11}$)
$E(X)=2\cdot\frac{6}{11}+3 \cdot \frac{1}{11} + 4 \cdot \frac{1}{11} +5\cdot \frac{3}{11}=\frac{34}{11}$ (Correct solution: $ \frac{45}{11}$)
Where is my mistake?
Should I start with $P(0)$ or $P(1)$?
 [/INST] The values that you got for the probabilities $P(X=x)$ are wrong. $P(2)=0$ and $P(3)=\frac6{11}$ and so on. If you correct this, then you get for the first one:\begin{align}P(X=3)&=P(X\le 3)-P(X<3)=F_X(3)=\frac6{11}\\[0.2cm]P(X=5)&=P(X\le 5)-P(X<5)=F_X(5)-F_X(5-)=\frac8{11}-\frac7{11}=\frac1{11}\end{align} Hence $P(A)=P(X=3)+P(X=5)=\frac6{11}+\frac1{11}=\frac7{11}$. 
For the expected value, note that from the given $F$ you can conlude that $$P(X=3)=\frac6{11},\;P(X=4)=\frac1{11},\;P(X=5)=\frac1{11},\;P(X=6)=\frac3{11}$$ Hence $$E[X]=3\cdot\frac6{11}+4\cdot\frac1{11}+5\cdot\frac1{11}+6\cdot\frac3{11}=\frac{45}{11}$$
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