91Ó°ÊÓ

To verify that \(f(x) = \frac{8}{7}\left(\frac{1}{2}\right)^{x}\) is a probability mass function, we need to ensure that the sum of \(f(x)\) over all possible values of \(x\) is equal to 1.Calculate the sum: \[\sum_{x=1}^{3} f(x) = \frac{8}{7}\left(\frac{1}{2}\right)^1 + \frac{8}{7}\left(\frac{1}{2}\right)^2 + \frac{8}{7}\left(\frac{1}{2}\right)^3 \]Calculate each term:\(\frac{8}{7} \times \frac{1}{2} = \frac{8}{14}\),\(\frac{8}{7} \times \frac{1}{4} = \frac{8}{28}\),\(\frac{8}{7} \times \frac{1}{8} = \frac{8}{56}\).Add these probabilities: \[\frac{8}{14} + \frac{8}{28} + \frac{8}{56} = \frac{32}{56} + \frac{16}{56} + \frac{8}{56} = \frac{56}{56} = 1\]The function \(f(x)\) satisfies the properties of a probability mass function.

To find \(P(X \leq 1)\), calculate the probability when \(x = 1\).\[P(X \leq 1) = f(1) = \frac{8}{7}\left(\frac{1}{2}\right)^1 = \frac{8}{14} = \frac{4}{7} \]

\(P(X > 1)\) requires adding up \(f(x)\) for \(x = 2\) and \(x = 3\).\[P(X > 1) = f(2) + f(3) = \frac{8}{7}\left(\frac{1}{2}\right)^2 + \frac{8}{7}\left(\frac{1}{2}\right)^3 \]\[= \frac{8}{28} + \frac{8}{56} = \frac{16}{56} + \frac{8}{56} = \frac{24}{56} = \frac{3}{7}\]

Since \(f(x)\) is only defined for \(x = 1, 2, 3\), \(P(2 < X < 6)\) means finding \(0\) since no values satisfy this.There are no \(x\) values in the range \(2 < x < 6\), so:\[P(2 < X < 6) = 0\]

This requires adding probabilities from Steps 2 and 3 since they cover disjoint sets:\[P(X \leq 1 \text{ or } X > 1) = P(X \leq 1) + P(X > 1)\]Using the results:\[= \frac{4}{7} + \frac{3}{7} = \frac{7}{7} = 1\]

All calculations are verified, and the function is a probability mass function. The probabilities requested are:\(P(X \leq 1) = \frac{4}{7}\),\(P(X > 1) = \frac{3}{7}\),\(P(2 < X < 6) = 0\),\(P(X \leq 1 \text{ or } X > 1) = 1\).