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Chapter 2: Problem 39

The random variable \(X\) has the following probability mass function $$ p(1)=\frac{1}{2}, \quad p(2)=\frac{1}{3}, \quad p(24)=\frac{1}{6} $$ Calculate \(E[X]\)

Short Answer

Expert verified
The expected value of the random variable \(X\) is \(E[X] = \frac{31}{6}\).

Step by step solution

01

Set up expected value formula with given pmf

Given the probability mass function (pmf), we can calculate the expected value using the formula: $$ E[X] = \sum_x xp(x) $$ In this case, the possible outcomes of the random variable \(X\) are \(1\), \(2\), and \(24\). We are also given their probabilities: \(p(1)=\frac{1}{2}\), \(p(2)=\frac{1}{3}\), and \(p(24)=\frac{1}{6}\). Plugging these probabilities into the expected value formula, we get: $$ E[X] = 1\cdot \frac{1}{2}+2\cdot \frac{1}{3}+24\cdot \frac{1}{6} $$
02

Calculate the expected value

Now we just need to compute the sum: $$ E[X] = 1\cdot \frac{1}{2}+2\cdot \frac{1}{3}+24\cdot \frac{1}{6} = \frac{1}{2}+\frac{2}{3}+4 $$ To make it easier to add the numbers together, we will first find a common denominator for the fractions. $$ E[X] = \frac{1}{2}+\frac{2}{3}+\frac{24}{6} = \frac{3}{6}+\frac{4}{6}+\frac{24}{6} $$ Now, we can add the fractions: $$ E[X] = \frac{3+4+24}{6} = \frac{31}{6} $$ So, the expected value of the random variable \(X\) is: $$ E[X] = \frac{31}{6} $$

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