A random variable $X$ has the probability distribution given in the following table.

$x$	1	4	6	8
$\mathrm{P}(X = x)$	$a$	$b$	$c$	$d$

Given that $\mathrm{E}(X) = 4$, $\mathrm{Var}(X) = \frac{19}{4}$ and $\mathrm{P}(X < 4) = \mathrm{P}(X > 4)$, find the values of $a$, $b$, $c$ and $d$.

1. Express $b$ in terms of $a$ using the sum of probabilities and the given probability condition.

The sum of all probabilities in a probability distribution is $1$:
$a + b + c + d = 1$
Using the condition $\mathrm{P}(X < 4) = \mathrm{P}(X > 4)$:
$a = c + d$
Substituting $c + d = a$ into the sum equation:
$a + b + a = 1 \implies 2a + b = 1 \implies b = 1 - 2a$

2. Use the expectation $\mathrm{E}(X) = 4$ to show that $c = d$.

The expectation is given by $\mathrm{E}(X) = \sum x \mathrm{P}(X=x)$:
$1(a) + 4(b) + 6(c) + 8(d) = 4$
Substitute $b = 1 - 2a$:
$a + 4(1 - 2a) + 6c + 8d = 4$
$a + 4 - 8a + 6c + 8d = 4 \implies -7a + 6c + 8d = 0$
Substitute $a = c + d$:
$-7(c + d) + 6c + 8d = 0$
$-7c - 7d + 6c + 8d = 0 \implies -c + d = 0 \implies c = d$

3. Express $a$ and $b$ entirely in terms of $c$.

Since $c = d$, we substitute $d$ for $c$ in our earlier expressions:
$a = c + d \implies a = 2c$
$b = 1 - 2a \implies b = 1 - 2(2c) = 1 - 4c$

4. Use the variance formula to solve for $c$.

The variance is given by $\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2$:
$\mathrm{E}(X^2) = 1^2(a) + 4^2(b) + 6^2(c) + 8^2(d) = a + 16b + 36c + 64d$
$a + 16b + 36c + 64d - 4^2 = \frac{19}{4}$
Substitute $a = 2c$, $b = 1 - 4c$, and $d = c$:
$2c + 16(1 - 4c) + 36c + 64(c) - 16 = \frac{19}{4}$
$2c + 16 - 64c + 36c + 64c - 16 = \frac{19}{4}$
$38c = \frac{19}{4} \implies c = \frac{19}{4 \times 38} = \frac{1}{8}$

5. Calculate the final values for $a$, $b$, and $d$.

$a = 2c = 2\left(\frac{1}{8}\right) = \frac{1}{4}$
$b = 1 - 4c = 1 - 4\left(\frac{1}{8}\right) = \frac{1}{2}$
$d = c = \frac{1}{8}$