1. **Problem:** An urn contains 10 balls: four '4's, three '3's, two '2's, and one '1'. Two balls are drawn without replacement. Find the probability that the first ball is '1' given the largest number drawn is '3'. 2. **Formula and rules:** Use conditional probability: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ where - $A$: first ball is '1' - $B$: largest number is '3' 3. **Calculate $P(B)$:** Largest number is '3' means no '4' drawn. Total ways to choose 2 balls: $$\binom{10}{2} = 45$$ Balls excluding '4's: 6 balls (3 '3's, 2 '2's, 1 '1') Ways to choose 2 balls with largest 3: $$\binom{6}{2} = 15$$ So, $$P(B) = \frac{15}{45} = \frac{1}{3}$$ 4. **Calculate $P(A \cap B)$:** First ball is '1' and largest is '3'. First ball '1' is fixed. Second ball must be '3' or '2' (to keep largest 3). Number of ways: 1 (ball '1') × 5 (balls '3' or '2') = 5 Total ways to pick 2 balls: 45 So, $$P(A \cap B) = \frac{5}{45} = \frac{1}{9}$$ 5. **Calculate conditional probability:** $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{9} \times 3 = \frac{1}{3} = 0.333$$ **Final answer:** 0.333 --- 2. **Problem:** Given $X \sim Binomial(n=20, p)$ and $$Var(X) = \frac{44}{5} - 10p$$ find $p$. 3. **Formula:** Variance of binomial: $$Var(X) = np(1-p)$$ 4. **Set equation:** $$20p(1-p) = \frac{44}{5} - 10p$$ 5. **Simplify:** $$20p - 20p^2 = \frac{44}{5} - 10p$$ 6. **Bring all terms to one side:** $$20p - 20p^2 + 10p - \frac{44}{5} = 0$$ $$30p - 20p^2 - \frac{44}{5} = 0$$ 7. **Multiply both sides by 5 to clear fraction:** $$5 \times (30p - 20p^2) - 44 = 0$$ $$150p - 100p^2 - 44 = 0$$ 8. **Rewrite:** $$-100p^2 + 150p - 44 = 0$$ 9. **Divide by -2 for simpler coefficients:** $$50p^2 - 75p + 22 = 0$$ 10. **Use quadratic formula:** $$p = \frac{75 \pm \sqrt{(-75)^2 - 4 \times 50 \times 22}}{2 \times 50}$$ $$= \frac{75 \pm \sqrt{5625 - 4400}}{100} = \frac{75 \pm \sqrt{1225}}{100}$$ $$= \frac{75 \pm 35}{100}$$ 11. **Two solutions:** $$p_1 = \frac{75 + 35}{100} = \frac{110}{100} = 1.1$$ (not valid, $p \leq 1$) $$p_2 = \frac{75 - 35}{100} = \frac{40}{100} = 0.4$$ **Final answer:** $p = 0.4$ --- 3. **Problem:** Number of trains arriving in 10 minutes is Poisson with mean $\lambda=3$. Find probability that number arriving in 20 minutes is less than 4. 4. **Formula:** For Poisson, number in 20 minutes has mean $$\mu = 2 \times 3 = 6$$ Probability $$P(X < 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$$ where $$P(X=k) = \frac{e^{-\mu} \mu^k}{k!}$$ 5. **Calculate:** $$P(X=0) = e^{-6} \frac{6^0}{0!} = e^{-6}$$ $$P(X=1) = e^{-6} \frac{6^1}{1!} = 6e^{-6}$$ $$P(X=2) = e^{-6} \frac{6^2}{2!} = 18e^{-6}$$ $$P(X=3) = e^{-6} \frac{6^3}{3!} = 36e^{-6}$$ 6. **Sum:** $$P(X<4) = e^{-6}(1 + 6 + 18 + 36) = e^{-6} \times 61$$ 7. **Numerical value:** $$e^{-6} \approx 0.00247875$$ $$P(X<4) \approx 0.00247875 \times 61 = 0.151$$ **Final answer:** 0.151