Question

If a discrete random variable X has the following probability distribution:

X	\(\frac{2}{3}\)	1	\(\frac{4}{3}\)
P(X)	\(c^2\)	\(c^2\)	c

then c is:

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{1}{5}\)

Answer 1

The Correct Option is A

To find the value of \( c \), we utilize the property that the sum of probabilities in a probability distribution equals 1. Given the probability distribution:

X	\(\frac{2}{3}\)	1	\(\frac{4}{3}\)
P(X)	\(c^2\)	\(c^2\)	c

set up the equation:

\[c^2 + c^2 + c = 1\]

Simplify the equation:

\[2c^2 + c = 1\]

Rearrange the equation into standard quadratic form:

\[2c^2 + c - 1 = 0\]

Solve this quadratic equation using the quadratic formula where \(a = 2\), \(b = 1\), and \(c = -1\):

\[c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Substitute the values:

\[c = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}\]

\[c = \frac{-1 \pm \sqrt{1 + 8}}{4}\]

\[c = \frac{-1 \pm \sqrt{9}}{4}\]

\[c = \frac{-1 \pm 3}{4}\]

This gives two potential solutions:

\[c = \frac{2}{4} = \frac{1}{2}\]

or

\[c = \frac{-4}{4} = -1\]

Since probabilities must be positive, \( c = -1 \) is not valid. Thus, the solution is:

\[c = \frac{1}{2}\]