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In the realm of probability, verifying a probability mass function (PMF) is crucial to ensure that it adheres to fundamental probability laws. A function qualifies as a PMF if it satisfies two essential conditions:

• Non-negativity: The function's value must be non-negative for every conceivable outcome.
• Normalization: The sum of all the probabilities of the possible outcomes must be exactly 1.

Consider the function given in this exercise: \(f(x) = \frac{2x+1}{25}\) for \(x = 0, 1, 2, 3, 4\). To verify its legitimacy as a PMF, we initially calculate \(f(x)\) for each value of \(x\). Here’s the breakdown:

• \(f(0) = \frac{1}{25}\)
• \(f(1) = \frac{3}{25}\)
• \(f(2) = \frac{5}{25}\)
• \(f(3) = \frac{7}{25}\)
• \(f(4) = \frac{9}{25}\)

Each computed \(f(x)\) meets the non-negativity condition. Adding them up, \(\frac{1}{25} + \frac{3}{25} + \frac{5}{25} + \frac{7}{25} + \frac{9}{25} = 1\), validates the normalization condition. Hence, our function qualifies as a valid probability mass function.

Probability distributions describe how probabilities are allocated across various possible outcomes. When dealing with discrete random variables, we commonly utilize discrete probability distributions, which specifically cater to countable, distinct outcomes.
A probability mass function (PMF) is used to illustrate the distribution. In this exercise, \(f(x)\) represents a discrete distribution over the sample set \(\{0, 1, 2, 3, 4\}\). The function assigns probability values respectively as probabilities to these discrete points.

• Each value of \(x\) has its own unique probability.
• The PMF provides us the probabilities for discrete outcomes directly.

For example, the provided probabilities for this discrete probability distribution are:

• \(P(X=0) = \frac{1}{25}\)
• \(P(X=1) = \frac{3}{25}\)
• \(P(X=2) = \frac{5}{25}\)
• \(P(X=3) = \frac{7}{25}\)
• \(P(X=4) = \frac{9}{25}\)

These values allow us to calculate more complex probabilities, such as cumulative or conditional probabilities, by further manipulating the PMF.

Once we have a verified probability mass function, calculating relevant probabilities becomes straightforward. Each scenario involves determining the probability for a specific outcome or range of outcomes based on the PMF.Calculate the probability for each requested scenario:

• For \(P(X=4)\): The PMF directly gives \(P(X=4) = f(4) = \frac{9}{25}\).
• To find \(P(X \leq 1)\): Sum probabilities from \(X=0\) to \(X=1\): \(P(X \leq 1) = \frac{1}{25} + \frac{3}{25} = \frac{4}{25}\).
• For \(P(2 \leq X < 4)\): Include probabilities where \(X=2\) and \(X=3\): \(P(2 \leq X < 4) = \frac{5}{25} + \frac{7}{25} = \frac{12}{25}\).
• Computing \(P(X > -10)\): Since all \(x\) values are greater than \(-10\), the probability sums to 1, or \(P(X > -10) = 1\).

These calculations demonstrate applying the PMF to derive explicit probabilities from given criteria, offering clarity on the possible outcomes’ likelihoods.