Solving for K in a Discrete Probability Distribution

A discrete random variable X has a probability distribution defined by P(X=x) for each possible value x. A fundamental property of any probability distribution is that the sum of the probabilities for all possible values of the random variable must equal 1. This means \(\sum P(X=x) = 1\).

In this given problem, the discrete random variable X can take values 1, 2, 3, 4, 5, 6, and 7. The corresponding probabilities are given in terms of K:

• P(X=1) = K
• P(X=2) = 2K
• P(X=3) = 2K
• P(X=4) = 3K
• P(X=5) = K²
• P(X=6) = 2K²
• P(X=7) = 7K² + K

To find the value of K, we use the property that the sum of all these probabilities is 1:

\(\sum_{x=1}^{7} P(X=x) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 1\)

Substitute the given probabilities into the equation:

\(K + 2K + 2K + 3K + K^2 + 2K^2 + 7K^2 + K = 1\)

Now, combine the terms involving K and the terms involving K²:

\((K + 2K + 2K + 3K + K) + (K^2 + 2K^2 + 7K^2) = 1\)

Summing the K terms:

\(K + 2K + 2K + 3K + K = (1+2+2+3+1)K = 9K\)

Summing the K² terms:

\(K^2 + 2K^2 + 7K^2 = (1+2+7)K^2 = 10K^2\)

So, the equation becomes:

\(9K + 10K^2 = 1\)

Rearrange this into a standard quadratic equation form (ax² + bx + c = 0):

\(10K^2 + 9K - 1 = 0\)

We can solve this quadratic equation for K. One common method is factoring. We look for two numbers that multiply to (10)(-1) = -10 and add up to 9. These numbers are +10 and -1.

Rewrite the middle term using these numbers:

\(10K^2 + 10K - K - 1 = 0\)

Group terms and factor by grouping:

\((10K^2 + 10K) - (K + 1) = 0\)

Factor out common terms from each group:

\(10K(K + 1) - 1(K + 1) = 0\)

Factor out the common binomial factor (K + 1):

\((10K - 1)(K + 1) = 0\)

This gives two possible solutions for K:

\(10K - 1 = 0 \implies 10K = 1 \implies K = \frac{1}{10}\)

\(K + 1 = 0 \implies K = -1\)

In a probability distribution, all probabilities P(X=x) must be non-negative (P(X=x) \(\ge\) 0). Let's check both possible values of K.

• If \(K = -1\): P(X=1) = K = -1. This is not a valid probability as probabilities cannot be negative. So, \(K = -1\) is rejected.
• If \(K = \frac{1}{10}\):
  • P(X=1) = \(K = \frac{1}{10}\) (\(\ge\) 0)
  • P(X=2) = \(2K = 2 \times \frac{1}{10} = \frac{2}{10}\) (\(\ge\) 0)
  • P(X=3) = \(2K = 2 \times \frac{1}{10} = \frac{2}{10}\) (\(\ge\) 0)
  • P(X=4) = \(3K = 3 \times \frac{1}{10} = \frac{3}{10}\) (\(\ge\) 0)
  • P(X=5) = \(K^2 = (\frac{1}{10})^2 = \frac{1}{100}\) (\(\ge\) 0)
  • P(X=6) = \(2K^2 = 2 \times (\frac{1}{10})^2 = \frac{2}{100}\) (\(\ge\) 0)
  • P(X=7) = \(7K^2 + K = 7 \times (\frac{1}{10})^2 + \frac{1}{10} = \frac{7}{100} + \frac{10}{100} = \frac{17}{100}\) (\(\ge\) 0)

All probabilities are non-negative when \(K = \frac{1}{10}\), and their sum is 1. Therefore, the valid value for K is \(\frac{1}{10}\).

Revision Table: Key Concepts

Additional Information: Properties of Probability Distributions

Understanding probability distributions is crucial in statistics. Here are some key properties and related concepts:

• Discrete vs. Continuous: Random variables can be discrete (countable outcomes) or continuous (outcomes over a range). The rules for their probability distributions differ.
• Probability Mass Function (PMF): For a discrete random variable, the probability distribution is often called a Probability Mass Function (PMF), denoted by P(X=x) or f(x). It gives the probability that the random variable X takes on a specific value x.
• Requirements for a Valid PMF: For a function P(x) to be a valid PMF, it must satisfy two conditions:
  • \(0 \le P(x) \le 1\) for all possible values of x.
  • \(\sum P(x) = 1\) over all possible values of x.
• Expected Value: The expected value (or mean) of a discrete random variable X is calculated as \(E(X) = \sum x \cdot P(X=x)\).
• Variance: The variance of a discrete random variable X is calculated as \(Var(X) = E(X^2) - [E(X)]^2\), where \(E(X^2) = \sum x^2 \cdot P(X=x)\).

Solving for unknown constants like K often involves using the sum of probabilities property, which frequently leads to solving algebraic equations, including quadratic equations as seen in this problem.