1. In an examination, the probability of a candidate solving a question in \( 1 / 2 \). Out of given 5 questions in the exam, what is the probability that the candidate was able to answer at least 2 questions? A. \( 1 / 64 \) B. \( 3 / 16 \) C. \( 1 / 2 \) D. \( 13 / 16 \)

To find the probability that a candidate answers at least 2 questions correctly out of 5, we can use the binomial probability formula. The probability of solving a question is \( p = \frac{1}{2} \) and \( q = \frac{1}{2} \) (not solving). The number of trials \( n = 5 \). We want to calculate the probabilities for 0 and 1 correct answers and subtract from 1. The probability that the candidate answers exactly \( k \) questions is given by: \[ P(X = k) = {n \choose k} p^k q^{n-k} \] So for \( k = 0 \) and \( k = 1 \): - For \( k = 0 \): \[ P(X = 0) = {5 \choose 0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^5 = 1 \cdot 1 \cdot \frac{1}{32} = \frac{1}{32} \] - For \( k = 1 \): \[ P(X = 1) = {5 \choose 1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^4 = 5 \cdot \frac{1}{2} \cdot \frac{1}{16} = \frac{5}{32} \] Adding these gives: \[ P(X \leq 1) = P(X = 0) + P(X = 1) = \frac{1}{32} + \frac{5}{32} = \frac{6}{32} = \frac{3}{16} \] Now, to find the probability of at least 2 questions solved: \[ P(X \geq 2) = 1 - P(X \leq 1) = 1 - \frac{3}{16} = \frac{16 - 3}{16} = \frac{13}{16} \] Thus, the answer is \( \text{D. } \frac{13}{16} \).