Question

The probability distribution of a random variable X is given by
\begin{tabular}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & $1 - 7a^2$ & $\frac{1}{2}a + \frac{1}{4}$ & $a^2$ \\ \hline \end{tabular}
If a > 0, then P(0 $<$ x $\le$ 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: Understanding the Concept:
For any probability distribution, the sum of all probabilities for all possible values of the random variable must be equal to 1. That is, \(\sum P(X=x_i) = 1\).
Step 2: Key Formula or Approach:
1. Set up an equation by summing the probabilities and equating to 1.
2. Solve the equation for the unknown parameter 'a'.
3. Use the value of 'a' to calculate the required probability.
Step 3: Detailed Explanation:
The sum of the probabilities is:
\[ P(X=0) + P(X=1) + P(X=2) = 1 \] \[ (1 - 7a^2) + \left(\frac{1}{2}a + \frac{1}{4}\right) + (a^2) = 1 \] Simplify the equation:
\[ 1 - 6a^2 + \frac{1}{2}a + \frac{1}{4} = 1 \] Subtract 1 from both sides:
\[ -6a^2 + \frac{1}{2}a + \frac{1}{4} = 0 \] To eliminate fractions, multiply the entire equation by 4:
\[ -24a^2 + 2a + 1 = 0 \] Multiply by -1 to make the leading coefficient positive:
\[ 24a^2 - 2a - 1 = 0 \] Solve this quadratic equation for 'a'. We can factor it:
\[ 24a^2 - 6a + 4a - 1 = 0 \] \[ 6a(4a - 1) + 1(4a - 1) = 0 \] \[ (6a + 1)(4a - 1) = 0 \] This gives two possible values for a: \(a = -\frac{1}{6}\) or \(a = \frac{1}{4}\).
The problem states that \(a>0\), so we must choose \(a = \frac{1}{4}\).
Now, we need to find \(P(0<X \le 2)\). This is the sum of probabilities for X = 1 and X = 2.
\[ P(0<X \le 2) = P(X=1) + P(X=2) \] \[ P(0<X \le 2) = \left(\frac{1}{2}a + \frac{1}{4}\right) + (a^2) \] Substitute \(a = \frac{1}{4}\) into this expression:
\[ P(0<X \le 2) = \left(\frac{1}{2}\left(\frac{1}{4}\right) + \frac{1}{4}\right) + \left(\frac{1}{4}\right)^2 \] \[ = \left(\frac{1}{8} + \frac{1}{4}\right) + \frac{1}{16} \] \[ = \left(\frac{1}{8} + \frac{2}{8}\right) + \frac{1}{16} = \frac{3}{8} + \frac{1}{16} \] \[ = \frac{6}{16} + \frac{1}{16} = \frac{7}{16} \] Step 4: Final Answer:
The value of P(0<x $\le$ 2) is $\frac{7{16}$}.