1. **Problem Statement:** Given a binomial random variable $X \sim B(8, 0.2)$, calculate various probabilities. 2. **Formula:** The binomial probability mass function is $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $n=8$, $p=0.2$, and $k$ is the number of successes. 3. **Calculations:** **a i. Calculate $P(X=3)$:** $$P(X=3) = \binom{8}{3} (0.2)^3 (0.8)^5$$ Calculate the binomial coefficient: $$\binom{8}{3} = \frac{8!}{3!5!} = 56$$ Calculate powers: $$(0.2)^3 = 0.008$$ $$(0.8)^5 = 0.32768$$ Multiply all: $$P(X=3) = 56 \times 0.008 \times 0.32768 = 0.1468$$ **a ii. Given $P(X=4) = 0.046$ (verified):** $$P(X=4) = \binom{8}{4} (0.2)^4 (0.8)^4 = 0.046$$ **b i. Calculate $P(X \leq 3)$:** Given as $0.944$. **b ii. Calculate $P(X \leq 2)$:** Given as $0.797$. **c i. Calculate $P(X > 3)$:** $$P(X > 3) = 1 - P(X \leq 3) = 1 - 0.944 = 0.056$$ **c ii. Calculate $P(X > 4)$:** $$P(X > 4) = 1 - P(X \leq 4)$$ Calculate $P(X=4)$ from a ii: $0.046$ $$P(X \leq 4) = P(X \leq 3) + P(X=4) = 0.944 + 0.046 = 0.99$$ So, $$P(X > 4) = 1 - 0.99 = 0.01$$ **d i. Calculate $P(X < 5)$:** $$P(X < 5) = P(X \leq 4) = 0.99$$ Given as $0.9816$ (approximate). **d ii. Calculate $P(X < 3)$:** $$P(X < 3) = P(X \leq 2) = 0.797$$ **e i. Calculate $P(X \geq 3)$:** $$P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.797 = 0.203$$ **e ii. Calculate $P(X \geq 1)$:** $$P(X \geq 1) = 1 - P(X=0)$$ Calculate $P(X=0)$: $$P(X=0) = \binom{8}{0} (0.2)^0 (0.8)^8 = 1 \times 1 \times 0.16777 = 0.1678$$ So, $$P(X \geq 1) = 1 - 0.1678 = 0.8322$$ **f i. Calculate $P(3 < X \leq 6)$:** $$P(3 < X \leq 6) = P(X=4) + P(X=5) + P(X=6)$$ Calculate each term: $$P(X=4) = 0.046$$ $$P(X=5) = \binom{8}{5} (0.2)^5 (0.8)^3 = 56 \times 0.00032 \times 0.512 = 0.0092$$ $$P(X=6) = \binom{8}{6} (0.2)^6 (0.8)^2 = 28 \times 0.000064 \times 0.64 = 0.00115$$ Sum: $$0.046 + 0.0092 + 0.00115 = 0.05635$$ **Final answers:** - $P(X=3) = 0.1468$ - $P(X=4) = 0.046$ - $P(X \leq 3) = 0.944$ - $P(X \leq 2) = 0.797$ - $P(X > 3) = 0.056$ - $P(X > 4) = 0.01$ - $P(X < 5) = 0.99$ - $P(X < 3) = 0.797$ - $P(X \geq 3) = 0.203$ - $P(X \geq 1) = 0.8322$ - $P(3 < X \leq 6) = 0.05635$