Calculating Probability P(X > 4) for a Discrete Random Variable

The problem asks us to find the probability \(P(X > 4)\) for a discrete random variable \(X\) with a given Probability Mass Function (PMF). The PMF is defined as \(P(x) = \dfrac{1}{2^x}\) for \(x = 1, 2, 3, \dots\).

A discrete random variable can only take specific, separated values. In this case, the possible values for \(X\) are the positive integers starting from 1.

The probability \(P(X > 4)\) means the probability that the random variable \(X\) takes a value strictly greater than 4. Since \(X\) can only be \(1, 2, 3, \dots\), the values greater than 4 are \(5, 6, 7, \dots\).

We can calculate \(P(X > 4)\) in two main ways:

Method 1: Summing the Probabilities for X > 4

\(P(X > 4) = P(X=5) + P(X=6) + P(X=7) + \dots\)

Using the given PMF \(P(x) = \dfrac{1}{2^x}\), we substitute the values of \(x\):

• \(P(X=5) = \dfrac{1}{2^5} = \dfrac{1}{32}\)
• \(P(X=6) = \dfrac{1}{2^6} = \dfrac{1}{64}\)
• \(P(X=7) = \dfrac{1}{2^7} = \dfrac{1}{128}\)
• and so on...

So, \(P(X > 4) = \dfrac{1}{32} + \dfrac{1}{64} + \dfrac{1}{128} + \dots\)

This is an infinite geometric series with the first term \(a = \dfrac{1}{32}\) and the common ratio \(r = \dfrac{1/64}{1/32} = \dfrac{1}{64} \times 32 = \dfrac{32}{64} = \dfrac{1}{2}\).

The sum of an infinite geometric series is given by the formula \(S = \dfrac{a}{1-r}\), provided \(|r| < 1\). Here, \(|1/2| < 1\), so the formula is applicable.

\(P(X > 4) = \dfrac{\dfrac{1}{32}}{1 - \dfrac{1}{2}} = \dfrac{\dfrac{1}{32}}{\dfrac{1}{2}}\)

To divide by a fraction, we multiply by its reciprocal:

\(P(X > 4) = \dfrac{1}{32} \times 2 = \dfrac{2}{32} = \dfrac{1}{16}\).

Method 2: Using the Complement Rule

The total probability for all possible values of a random variable is 1. The event \(X > 4\) is the complement of the event \(X \le 4\).

So, \(P(X > 4) = 1 - P(X \le 4)\).

The event \(X \le 4\) means \(X\) takes values \(1, 2, 3, \text{ or } 4\).

\(P(X \le 4) = P(X=1) + P(X=2) + P(X=3) + P(X=4)\).

Using the PMF \(P(x) = \dfrac{1}{2^x}\):

• \(P(X=1) = \dfrac{1}{2^1} = \dfrac{1}{2}\)
• \(P(X=2) = \dfrac{1}{2^2} = \dfrac{1}{4}\)
• \(P(X=3) = \dfrac{1}{2^3} = \dfrac{1}{8}\)
• \(P(X=4) = \dfrac{1}{2^4} = \dfrac{1}{16}\)

Now, sum these probabilities:

\(P(X \le 4) = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16}\)

To add these fractions, find a common denominator, which is 16:

\(P(X \le 4) = \dfrac{8}{16} + \dfrac{4}{16} + \dfrac{2}{16} + \dfrac{1}{16} = \dfrac{8 + 4 + 2 + 1}{16} = \dfrac{15}{16}\).

Finally, use the complement rule:

\(P(X > 4) = 1 - P(X \le 4) = 1 - \dfrac{15}{16}\)

\(P(X > 4) = \dfrac{16}{16} - \dfrac{15}{16} = \dfrac{16 - 15}{16} = \dfrac{1}{16}\).

Both methods yield the same result. The value of \(P(X > 4)\) is \(\dfrac{1}{16}\).

Revision Table: Key Concepts

Additional Information on Discrete Probability

Understanding the Probability Mass Function (PMF) is fundamental for working with discrete random variables. The PMF, denoted by \(P(x)\) or \(f_X(x)\), must satisfy two conditions:

1. \(0 \le P(x) \le 1\) for all possible values of \(x\).
2. \(\sum_x P(x) = 1\), where the sum is taken over all possible values of \(x\).

In our problem, the PMF is \(P(x) = \dfrac{1}{2^x}\) for \(x = 1, 2, 3, \dots\). Let's quickly check if the sum of probabilities is 1:

\(\sum_{x=1}^{\infty} \dfrac{1}{2^x} = \dfrac{1}{2^1} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \dots = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dots\)

This is an infinite geometric series with \(a = \dfrac{1}{2}\) and \(r = \dfrac{1}{2}\).

The sum is \(\dfrac{a}{1-r} = \dfrac{\dfrac{1}{2}}{1 - \dfrac{1}{2}} = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2}} = 1\).

This confirms that the given function is a valid PMF for a discrete random variable. The problem of calculating \(P(X > 4)\) then becomes a calculation based on this valid probability distribution. Both the summation method (using the geometric series formula) and the complement rule method are standard techniques for solving such problems in probability theory for discrete distributions.