Understanding Probability Mass Functions (PMF)

A Probability Mass Function (PMF) describes the probability that a discrete random variable takes on a specific value. For a function to be a valid PMF, it must satisfy two main properties:

• The probability for each value of the random variable must be non-negative, i.e., \(P(X = x) \ge 0\) for all possible values of \(x\).
• The sum of all probabilities for all possible values of the random variable must be equal to 1, i.e., \(\sum P(X = x) = 1\).

In this question, we are given a table that represents a probability mass function for a discrete random variable X:

To find the value of the constant \(k\), we must use the property that the sum of all probabilities in a probability mass function must equal 1.

Calculating k using the PMF Property

We need to sum all the given probabilities and set the total equal to 1. The probabilities are given in terms of \(k\) for each value of \(X\).

Sum of probabilities = \(P(X = -2) + P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\)

So, we have the equation:

\(2k + 3k + 4k + 3k + 2k + k = 1\)

Now, let's combine the terms on the left side of the equation:

\((2 + 3 + 4 + 3 + 2 + 1)k = 1\)

\(15k = 1\)

To find the value of \(k\), we divide both sides of the equation by 15:

\(k = \frac{1}{15}\)

Thus, the value of \(k\) that makes the given function a valid probability mass function is \(\frac{1}{15}\).

Let's verify the probabilities with \(k = \frac{1}{15}\):

• \(P(X = -2) = 2k = 2 \times \frac{1}{15} = \frac{2}{15}\)
• \(P(X = -1) = 3k = 3 \times \frac{1}{15} = \frac{3}{15}\)
• \(P(X = 0) = 4k = 4 \times \frac{1}{15} = \frac{4}{15}\)
• \(P(X = 1) = 3k = 3 \times \frac{1}{15} = \frac{3}{15}\)
• \(P(X = 2) = 2k = 2 \times \frac{1}{15} = \frac{2}{15}\)
• \(P(X = 3) = k = \frac{1}{15}\)

All probabilities are non-negative. Let's sum them:

\(\frac{2}{15} + \frac{3}{15} + \frac{4}{15} + \frac{3}{15} + \frac{2}{15} + \frac{1}{15} = \frac{2 + 3 + 4 + 3 + 2 + 1}{15} = \frac{15}{15} = 1\)

Since the sum of probabilities is 1, the value \(k = \frac{1}{15}\) is correct.

Revision Table: Key Concepts for PMF Calculations

Additional Information: Discrete Random Variables

A discrete random variable is a variable whose value can only take on a finite number of values or a countably infinite number of values. Examples include the number of heads in a series of coin flips, the number of cars passing a point on a road in an hour, or the outcome when rolling a die.

The probability mass function (PMF) is specifically used for discrete random variables. For continuous random variables, we use a Probability Density Function (PDF) instead. While a PMF gives the probability of a specific outcome, a PDF gives the likelihood of an outcome within a range, and the probability is found by integrating the PDF over that range. The property that the total probability sums to 1 holds for both PMF (\(\sum P(X=x) = 1\)) and PDF (\(\int f(x) dx = 1\)).

Understanding how to find unknown constants in probability distributions is a fundamental skill in probability and statistics. It ensures that the probabilities are correctly scaled so that their sum or integral equals 1.