1. **Problem 3:** Given events A, B, and C with $$P(A) = \frac{1}{2}, \quad P(C) = \frac{1}{2}, \quad P(A \cup B) = \frac{5}{8}$$ and A and C are mutually exclusive. **a. Find** $P(A \cup C)$. 2. Since A and C are mutually exclusive, by the addition rule: $$P(A \cup C) = P(A) + P(C) = \frac{1}{2} + \frac{1}{2} = 1$$ 3. Given A and B are independent, show $P(B) = \frac{3}{8}$. 4. Independence means: $$P(A \cap B) = P(A)P(B)$$ 5. Using the formula for union: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Substitute values: $$\frac{5}{8} = \frac{1}{2} + P(B) - \left(\frac{1}{2} \times P(B)\right)$$ 6. Simplify: $$\frac{5}{8} = \frac{1}{2} + P(B) - \frac{P(B)}{2} = \frac{1}{2} + \frac{P(B)}{2}$$ 7. Subtract $\frac{1}{2}$ from both sides: $$\frac{5}{8} - \frac{1}{2} = \frac{P(B)}{2} \implies \frac{5}{8} - \frac{4}{8} = \frac{P(B)}{2} \implies \frac{1}{8} = \frac{P(B)}{2}$$ 8. Multiply both sides by 2: $$P(B) = \frac{2}{8} = \frac{1}{4}$$ **Note:** This contradicts the problem's claim that $P(B) = \frac{3}{8}$. Re-checking step 6: $$\frac{5}{8} = \frac{1}{2} + P(B) - \frac{1}{2}P(B) = \frac{1}{2} + P(B)\left(1 - \frac{1}{2}\right) = \frac{1}{2} + \frac{P(B)}{2}$$ So the calculation is correct, $P(B) = \frac{1}{4}$, not $\frac{3}{8}$. Possibly a typo in the question. 9. **b. Find** $P(A|B)$. By definition: $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)P(B)}{P(B)} = P(A) = \frac{1}{2}$$ --- 10. **Problem 4:** Discrete random variable $X$ with distribution: | $x$ | 1 | 2 | 3 | 4 | |---|---|---|---|---| | $P(X=x)$ | 0.3 | 0.2 | 0.35 | 0.15 | **i. Find** $F(3.5) = P(X \leq 3.5)$. 11. Sum probabilities for $x \leq 3$: $$F(3.5) = P(X=1) + P(X=2) + P(X=3) = 0.3 + 0.2 + 0.35 = 0.85$$ **ii. Find** $P(3X - 3 < X + 2)$. 12. Simplify inequality: $$3X - 3 < X + 2 \implies 3X - X < 2 + 3 \implies 2X < 5 \implies X < \frac{5}{2} = 2.5$$ 13. So, $$P(X < 2.5) = P(X=1) + P(X=2) = 0.3 + 0.2 = 0.5$$ **iii. Find** $E(X)$. 14. Expected value: $$E(X) = \sum x P(X=x) = 1(0.3) + 2(0.2) + 3(0.35) + 4(0.15) = 0.3 + 0.4 + 1.05 + 0.6 = 2.35$$ **iv. Show that** $Var(X) = 1.1275$. 15. Variance formula: $$Var(X) = E(X^2) - [E(X)]^2$$ Calculate $E(X^2)$: $$E(X^2) = 1^2(0.3) + 2^2(0.2) + 3^2(0.35) + 4^2(0.15) = 0.3 + 0.8 + 3.15 + 2.4 = 6.65$$ 16. Then: $$Var(X) = 6.65 - (2.35)^2 = 6.65 - 5.5225 = 1.1275$$ **v. Find** $Var(5 - 2X)$. 17. Use variance property: $$Var(aX + b) = a^2 Var(X)$$ Here, $a = -2$, $b = 5$: $$Var(5 - 2X) = (-2)^2 Var(X) = 4 \times 1.1275 = 4.51$$