[THOUGHT] We are given a discrete random variable $X$ with a probability mass function defined for $x=0,1,2,3$ and $x=4$. The probability for $x=4$ is $a$, and we need to compute $\text{V}\left(\dfrac{1}{a}X\right)$.

First, we find the value of $a$ by normalizing the probabilities. The probabilities for $x=0,1,2,3$ are:

• $x=0$: $P(X=0) = |2\cdot0 - 1|/12 = 1/12$
• $x=1$: $P(X=1) = |2\cdot1 - 1|/12 = 1/12$
• $x=2$: $P(X=2) = |4-1|/12 = 3/12 = 1/4$
• $x=3$: $P(X=3) = |6-1|/12 = 5/12$

The sum of these probabilities is: [ \frac{1}{12} + \frac{1}{12} + \frac{1}{4} + \frac{5}{12} = \frac{1 + 1 + 3 + 5}{12} = \frac{10}{12} = \frac{5}{6} ] Since the total probability must be 1, the probability for $x=4$ is: [ a = 1 - \frac{5}{6} = \frac{1}{6} ]

The variance of a scaled random variable is $\text{V}(cX) = c^2 \text{V}(X)$. Here, $c = 1/a$, so: [ \text{V}\left(\frac{X}{a}\right) = \left(\frac{1}{a}\right)^2 \text{V}(X) = \frac{1}{a^2} \text{V}(X) ] With $a = 1/6$, we have $a^2 = (1/6)^2 = 1/36$, so: [ \text{V}\left(\frac{X}{a}\right) = \frac{1}{1/36} \text{V}(X) = 36 \text{V}(X) ]

Next, we compute the variance of $X$ using $\text{V}(X) = E[X^2] - (E[X])^2$.

First, compute $E[X]$: [ E[X] = \sum x \cdot P(X=x) = 0 \cdot \frac{1}{12} + 1 \cdot \frac{1}{12} + 2 \cdot \frac{3}{12} + 3 \cdot \frac{5}{12} + 4 \cdot \frac{2}{12} ] [ = 0 + \frac{1}{12} + \frac{6}{12} + \frac{15}{12} + \frac{8}{12} = \frac{1+6+15+8}{12} = \frac{30}{12} = \frac{5}{2} ]

Next, compute $E[X^2]$: [ E[X^2] = \sum x^2 \cdot P(X=x) = 0^2 \cdot \frac{1}{12} + 1^2 \cdot \frac{1}{12} + 2^2 \cdot \frac{3}{12} + 3^2 \cdot \frac{5}{12} + 4^2 \cdot \frac{2}{12} ] [ = 0 + \frac{1}{12} + \frac{12}{12} + \frac{45}{12} + \frac{32}{12} = \frac{1+12+45+32}{12} = \frac{90}{12} = \frac{15}{2} ]

Now, compute $\text{V}(X)$: [ \text{V}(X) = E[X^2] - (E[X])^2 = \frac{15}{2} - \left(\frac{5}{2}\right)^2 = \frac{15}{2} - \frac{25}{4} = \frac{30}{4} - \frac{25}{4} = \frac{5}{4} ]

Finally, substitute into the variance formula: [ \text{V}\left(\frac{X}{a}\right) = 36 \cdot \frac{5}{4} = 36 \cdot \frac{5}{4} = 45 ]

The answer is 45. \boxed{45}

[THOUGHT] I will use Python to compute the variance of (X/a) by first finding the value of a and then the moments of X.

[PYTHON]

# Define the probabilities for x=0,1,2,3
p0 = 1/12
p1 = 1/12
p2 = 3/12
p3 = 5/12

# The probability for x=4 is a, which we compute by normalization
total_prob = p0 + p1 + p2 + p3
a = 1 - total_prob  # which should be 1/6

# Now, compute E[X] and E[X^2]
x_values = [0, 1, 2, 3, 4]
probabilities = [p0, p1, p2, p3, a]

# Expected value E[X]
ex = sum(x * prob for x, prob in zip(x_values, probabilities))

# Expected value of X squared
ex2 = sum(x*x * prob for x, prob in zip(x_values, probabilities))

# Variance of X
var_x = ex2 - ex**2

# Variance of (X/a)
var_scaled = (1/a)**2 * var_x

print(f"V(X) = {var_x}")
print(f"V(X/a) = {var_scaled}")

[/PYTHON]

[THOUGHT] I will use Python to compute the variance of (X/a) by first finding the value of a and then the moments of X.

[PYTHON] from sympy import symbols, sum

Define the random variable values

x_values = [0, 1, 2, 3, 4]

Define the probabilities for x=0,1,2,3

p0 = 1/12 p1 = 1/12 p2 = 3/12 p3 = 5/12

Total probability for x=0,1,2,3

total_prob = p0 + p1 + p2 + p3

Probability for x=4 (a)

a = 1 - total_prob

Expected value E[X]

ex = sum(x * prob for x, prob in zip(x_values, [p0, p1, p2, p3, a]))

Expected value of X squared E[X^2]

ex2 = sum(x*x * prob for x, prob in zip(x_values, [p0, p1, p2, p3, a]))

Variance of X

var_x = ex2 - ex**2

Variance of (X/a)

var_scaled = (1/a)**2 * var_x

print(f"V(X) = {var_x}") print(f"V(X/a) = {var_scaled}") [/PYTHON]

\boxed{45}