A. Let \( f(x)=\left\{\begin{array}{lr}\frac{2^{6-x}}{64} & x=1,2,3,4,5 \\ \frac{1}{64} & x=6,7\end{array}\right. \) Hence, obtain the \( E(x) \)

To find the expected value \( E(x) \) of the given function \( f(x) \), we first need to determine the probabilities of each value of \( x \). The probabilities for \( x = 1, 2, 3, 4, 5 \) are \( \frac{2^{6-x}}{64} \), and for \( x = 6, 7 \), it’s \( \frac{1}{64} \). The expected value \( E(x) \) is calculated using the formula: \[ E(x) = \sum_{x=1}^{7} x \cdot P(x) \] Calculating the contributions: - For \( x = 1, 2, 3, 4, 5 \): \[ P(1) = \frac{2^{5}}{64}, \, P(2) = \frac{2^{4}}{64}, \, P(3) = \frac{2^{3}}{64}, \, P(4) = \frac{2^{2}}{64}, \, P(5) = \frac{2^{1}}{64} \] Calculating these gives us \( \frac{32}{64}, \frac{16}{64}, \frac{8}{64}, \frac{4}{64}, \frac{2}{64} \) respectively. - For \( x = 6 \) and \( x = 7 \): \[ P(6) = P(7) = \frac{1}{64} \] Substituting these into the expected value equation, compute: \[ E(x) = 1 \cdot \frac{32}{64} + 2 \cdot \frac{16}{64} + 3 \cdot \frac{8}{64} + 4 \cdot \frac{4}{64} + 5 \cdot \frac{2}{64} + 6 \cdot \frac{1}{64} + 7 \cdot \frac{1}{64} \] Calculating the sum: \[ = \frac{32 + 32 + 24 + 16 + 10 + 6 + 7}{64} = \frac{127}{64} \] Thus, the expected value \( E(x) \) is \( \frac{127}{64} \). Enjoy your math exploration, and remember: breaking complex problems into simple parts makes them easier to solve!