3 Given that \( Y \sim \mathrm{~B}(11,0.23) \), calculate: \( \begin{array}{ll}\text { a } \mathrm{P}(Y \neq 3) & \text { b } \mathrm{P}[Y<\mathrm{E}(Y)]\end{array} \)

To solve the problems related to the binomial distribution \( Y \sim \mathrm{B}(11,0.23) \): **a) To find \( P(Y \neq 3) \):** First, we calculate \( P(Y = 3) \): \[ P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( n = 11 \), \( k = 3 \), and \( p = 0.23 \). \[ P(Y = 3) = \binom{11}{3} (0.23)^3 (0.77)^{11-3} \] Calculating the components step-by-step: - Calculate \( \binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165 \). - \( (0.23)^3 \approx 0.012167 \). - \( (0.77)^8 \approx 0.097656 \). Now plug these values into the equation: \[ P(Y = 3) \approx 165 \times 0.012167 \times 0.097656 \approx 0.1931 \] Thus, \[ P(Y \neq 3) = 1 - P(Y = 3) \approx 1 - 0.1931 \approx 0.8069. \] **b) To find \( P(Y < E(Y)) \):** First, calculate the expected value \( E(Y) \): \[ E(Y) = n \cdot p = 11 \cdot 0.23 = 2.53. \] Since \( Y \) must be an integer, we look for \( P(Y < 2.53) \), which is equivalent to \( P(Y \leq 2) \). Now, calculate \( P(Y \leq 2) \): \[ P(Y = 0) = \binom{11}{0} (0.23)^0 (0.77)^{11} \approx 0.0777, \] \[ P(Y = 1) = \binom{11}{1} (0.23)^1 (0.77)^{10} \approx 0.1929, \] \[ P(Y = 2) = \binom{11}{2} (0.23)^2 (0.77)^9 \approx 0.2269. \] Adding these probabilities together: \[ P(Y \leq 2) \approx 0.0777 + 0.1929 + 0.2269 \approx 0.4975. \] Final answers: \[ \text{a) } P(Y \neq 3) \approx 0.8069, \quad \text{b) } P(Y < 2.53) \approx 0.4975. \]