Question

The probability distribution of a random variable X is given by

X	0	1	2
P(X)	\(1 - 7a^2\)	\(\tfrac{1}{2}a + \tfrac{1}{4}\)	\(a^2\)

If \(a > 0\), then \(P(0 < X \leq 2)\) is equal to

• A. $\frac{1}{16}$
• B. $\frac{3}{18}$
• C. $\frac{7}{16}$
• D. $\frac{9}{16}$

Hint:

The two fundamental properties of a discrete probability distribution are: 1) \(0 \le P(X=x_i) \le 1\) for all \(x_i\), and 2) \(\sum P(X=x_i) = 1\). The second property is almost always the starting point for finding unknown parameters.

Answer 1

The Correct Option is C

Step 1: Concept Foundation: For any probability distribution, the sum of probabilities across all possible random variable values must equal 1, represented as \(\sum P(X=x_i) = 1\).

Step 2: Methodology:
1. Formulate an equation by summing probabilities and setting the total to 1.
2. Solve for the unknown parameter 'a'.
3. Utilize the determined value of 'a' to compute the desired probability.

Step 3: Calculation Breakdown:
The sum of probabilities is:
\[ P(X=0) + P(X=1) + P(X=2) = 1 \]
Substituting the given expressions:
\[ (1 - 7a^2) + \left(\frac{1}{2}a + \frac{1}{4}\right) + (a^2) = 1 \]
Simplifying the equation:
\[ 1 - 6a^2 + \frac{1}{2}a + \frac{1}{4} = 1 \]
Subtracting 1 from both sides yields:
\[ -6a^2 + \frac{1}{2}a + \frac{1}{4} = 0 \]
Multiplying by 4 to clear fractions:
\[ -24a^2 + 2a + 1 = 0 \]
Multiplying by -1 to achieve a positive leading coefficient:
\[ 24a^2 - 2a - 1 = 0 \]
Factoring the quadratic equation for 'a':
\[ 24a^2 - 6a + 4a - 1 = 0 \]
\[ 6a(4a - 1) + 1(4a - 1) = 0 \]
\[ (6a + 1)(4a - 1) = 0 \]
The potential values for 'a' are \(a = -\frac{1}{6}\) or \(a = \frac{1}{4}\). Given the condition \(a>0\), we select \(a = \frac{1}{4}\).
The objective is to find \(P(0<X \le 2)\), which is \(P(X=1) + P(X=2)\).
\[ P(0<X \le 2) = P(X=1) + P(X=2) \]
\[ P(0<X \le 2) = \left(\frac{1}{2}a + \frac{1}{4}\right) + (a^2) \]
Substituting \(a = \frac{1}{4}\):
\[ P(0<X \le 2) = \left(\frac{1}{2}\left(\frac{1}{4}\right) + \frac{1}{4}\right) + \left(\frac{1}{4}\right)^2 \]
\[ = \left(\frac{1}{8} + \frac{1}{4}\right) + \frac{1}{16} \]
\[ = \left(\frac{1}{8} + \frac{2}{8}\right) + \frac{1}{16} = \frac{3}{8} + \frac{1}{16} \]
\[ = \frac{6}{16} + \frac{1}{16} = \frac{7}{16} \]

Step 4: Conclusion:
The calculated value for P(0<x $\le$ 2) is $\frac{7}{16}$.