Question

The range of a discrete random variable X is\( \{1, 2, 3\}\) and the probabilities of its elements are given by \(P(X=1) = 3k^3\), \(P(X=2) = 2k^2\) and \(P(X=3) = 7-19k\). Then \(P(X=3) = \)}

• A. \( \frac{2}{3} \)
• B. \( \frac{2}{9} \)
• C. \( \frac{1}{9} \)
• D. \( \frac{4}{9} \)

Hint:

The two fundamental rules for a probability distribution are: 1) The probability of any individual outcome must be between 0 and 1, inclusive. 2) The sum of the probabilities of all possible outcomes must equal 1. Always check your derived constants against both rules.

Answer 1

The Correct Option is A

For any discrete random variable, the sum of the probabilities of all possible outcomes must be equal to 1.
So, \( P(X=1) + P(X=2) + P(X=3) = 1 \).
Substitute the given expressions in terms of k:
\( 3k^3 + 2k^2 + (7-19k) = 1 \).
Rearrange this into a cubic equation:
\( 3k^3 + 2k^2 - 19k + 6 = 0 \).
We need to find a root of this equation. By the Rational Root Theorem, we can test integer factors of 6.
Let's test \( k=2 \): \( 3(2)^3 + 2(2)^2 - 19(2) + 6 = 3(8) + 2(4) - 38 + 6 = 24 + 8 - 38 + 6 = 38 - 38 = 0 \).
So, \( k=2 \) is a root.
Let's check if this value of k is valid. Probabilities must be between 0 and 1.
If \( k=2 \), \( P(X=1) = 3(2)^3 = 24 \), which is greater than 1. So, \( k=2 \) is not a valid solution.
Let's test \( k=1/3 \): \( 3(1/3)^3 + 2(1/3)^2 - 19(1/3) + 6 = 3/27 + 2/9 - 19/3 + 6 = 1/9 + 2/9 - 57/9 + 54/9 = (1+2-57+54)/9 = 0 \).
So, \( k=1/3 \) is another root. Let's check if it's valid.
\( P(X=1) = 3(1/3)^3 = 3/27 = 1/9 \). (Valid)
\( P(X=2) = 2(1/3)^2 = 2/9 \). (Valid)
\( P(X=3) = 7 - 19(1/3) = 7 - 19/3 = (21-19)/3 = 2/3 \). (Valid)
Since \( k=1/3 \) gives a valid probability distribution, this is the correct value of k.
The question asks for the value of \( P(X=3) \).
\( P(X=3) = \frac{2}{3} \).