The probability distribution of a discrete random variable $X$ is given as follows:

$x$	$-2$	$-1$	$0$	$1$	$3$
$\text{P}(X = x)$	$0.1$	$a$	$0.3$	$b$	$c$

Given that $\text{E}(X) = 0.9$ and $\text{Var}(X) = 2.99$, find the values of $a$, $b$ and $c$.

1. Formulate an equation using the property that the sum of all probabilities in a probability distribution is 1.

$\sum \text{P}(X=x) = 1$
$0.1 + a + 0.3 + b + c = 1$
$a + b + c = 0.6 \quad \text{--- (1)}$

2. Formulate a second equation using the given expectation $\text{E}(X)$.

$\text{E}(X) = 0.9$
$-2(0.1) + (-1)(a) + 0(0.3) + 1(b) + 3(c) = 0.9$
$-0.2 - a + b + 3c = 0.9$
$-a + b + 3c = 1.1 \quad \text{--- (2)}$

3. Formulate a third equation using the given variance $\text{Var}(X)$ and the relationship $\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$.

$\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$
$2.99 = \text{E}(X^2) - (0.9)^2$
$2.99 = \text{E}(X^2) - 0.81$
$\text{E}(X^2) = 3.8$

$(-2)^2(0.1) + (-1)^2(a) + 0^2(0.3) + 1^2(b) + 3^2(c) = 3.8$
$4(0.1) + a + b + 9c = 3.8$
$0.4 + a + b + 9c = 3.8$
$a + b + 9c = 3.4 \quad \text{--- (3)}$

4. Solve the system of three linear equations simultaneously to find $a$, $b$, and $c$.

Subtracting (1) from (3):
$(a + b + 9c) - (a + b + c) = 3.4 - 0.6$
$8c = 2.8 \implies c = 0.35$

Substituting $c = 0.35$ into (1) and (2):
$a + b + 0.35 = 0.6 \implies a + b = 0.25 \quad \text{--- (4)}$
$-a + b + 3(0.35) = 1.1 \implies -a + b = 0.05 \quad \text{--- (5)}$

Adding (4) and (5):
$(a + b) + (-a + b) = 0.25 + 0.05$
$2b = 0.30 \implies b = 0.15$

Substituting $b = 0.15$ into (4):
$a + 0.15 = 0.25 \implies a = 0.1$