Exercícios - Variáveis Aleatórias e Distrib de Prob - 2024-1

Exercícios quinta-feira, 16 de maio de 2024 01:03 Página 1 de probability models Página 2 de probability models Página 3 de probability models Página 4 de probability models Página 5 de probability models Página 6 de probability models 4) Passo 1 (a) Let X be the event of getting maximum value in two rolls. The values of the outcomes are \{1, 2, 3, 4, 5, 6\}. Passo 2 (b) Let X be the event of getting the minimum value in two rolls.The values of the outcomes would be \{1, 2, 3, 4, 5, 6\}, for example if we rolled 1 and 2, the minimum value is 1, so for the minimum value to appear in two roll is as said above, \{1, 2, 3, 4, 5, 6\}. Passo 3 (c) To find the sum of two rolls; \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \} Therefore the sum of two rolls are, \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}. Passo 4 (d) Let X the value of the difference of first and second roll: Say we rolled a 1 on the first roll and 6 for the second. X = 1 - 6 = -5 For the 2 on the first roll and 6 on the second, we got -4, repeat the sequence and you will get, \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\} 5) Passo 1 (a) P(X = 1) = P(\{(1, 1)\}) = \frac{1}{36} P(X = 2) = P(\{(2, 1)(2, 2)(2, 1)\}) = \frac{1}{12} P(X = 3) = P(\{(3, 1)(3, 2)(3, 3)(2, 3)(1, 3)\}) = \frac{5}{36} P(X = 4) = P(\{(4, 1)(4, 2)(4, 3)(4, 4)(3, 4)(2, 4)(1, 4)\}) = \frac{7}{36} P(X = 5) = P(\{(5, 1)(5, 2)(5, 3)(5, 4)(5, 5)(4, 5)(3, 5)(2, 5)(1, 5)\}) = \frac{1}{4} P(X = 6) = P(\{(6, 1)(6, 2)(6, 3)(6, 4)(6, 5)(6, 6)(5, 6)(4, 6)(3, 6)(2, 6)(1, 6)\}) = \frac{11}{36} Passo 2 (b) P(X = 6) = P(\{(6, 1)(6, 2)(6, 3)(6, 4)(6, 5)(6, 6)(5, 6)(4, 6)(3, 6)(2, 6)(1, 6)\}) = \frac{11}{36} P(X = 5) = P(\{(5, 1)(5, 2)(5, 3)(5, 4)(5, 5)(4, 5)(3, 5)(2, 5)(1, 5)\}) = \frac{1}{4} P(X = 4) = P(\{(4, 1)(4, 2)(4, 3)(4, 4)(3, 4)(2, 4)(1, 4)\}) = \frac{7}{36} P(X = 3) = P(\{(3, 1)(3, 2)(3, 3)(2, 3)(1, 3)\}) = \frac{5}{36} P(X = 2) = P(\{(2, 1)(2, 2)(2, 1)\}) = \frac{1}{12} P(X = 1) = P(\{(1, 1)\}) = \frac{1}{36} (c) P(X = 2) = P(\{(1, 1)\}) = \frac{1}{36} P(X = 3) = P(\{(1, 2)(2, 1)\}) = \frac{1}{18} P(X = 4) = P(\{(1, 3)(2, 2)(3, 1)\}) = \frac{1}{12} P(X = 5) = P(\{(1, 4)(2, 3)(3, 2)(4, 1)\}) = \frac{1}{9} P(X = 6) = P(\{(1, 5)(2, 4)(3, 3)(4, 2)(5, 1)\}) = \frac{5}{36} P(X = 7) = P(\{(1, 6)(2, 5)(3, 4)(4, 3)(5, 2)(6, 1)\}) = \frac{1}{6} P(X = 8) = P(\{(2, 6)(3, 5)(4, 4)(5, 3)(6, 2)\}) = \frac{5}{36} P(X = 9) = P(\{(3, 6)(4, 5)(5, 4)(6, 3)\}) = \frac{1}{9} P(X = 10) = P(\{(4, 6)(5, 5)(6, 4)\}) = \frac{1}{12} P(X = 11) = P(\{(5, 6)(6, 5)\}) = \frac{1}{18} P(X = 12) = P(\{(6, 6)\}) = \frac{1}{36} Passo 4 (d) P(X = 5) = P(\{(6, 1)\}) = \frac{1}{36} P(X = 4) = P(\{(6, 2)(5, 1)\}) = \frac{1}{18} P(X = 3) = P(\{(6, 3)(5, 2)(4, 1)\}) = \frac{1}{12} P(X = 2) = P(\{(6, 4)(5, 3)(4, 2)(3, 1)\}) = \frac{1}{9} P(X = 1) = P(\{(6, 5)(5, 4)(4, 3)(3, 2)(2, 1)\}) = \frac{5}{36} P(X = 0) = P(\{(6, 6)(5, 5)(4, 4)(3, 3)(2, 2)(1, 1)\}) = \frac{1}{6} P(X = -1) = P(\{(5, 6)(4, 5)(3, 4)(2, 3)(1, 2)\}) = \frac{5}{36} P(X = -2) = P(\{(4, 6)(3, 5)(2, 4)(1, 3)\}) = \frac{1}{9} P(X = -3) = P(\{(3, 6)(2, 5)(1, 4)\}) = \frac{1}{12} P(X = -4) = P(\{(2, 6)(1, 5)\}) = \frac{1}{18} P(X = -5) = P(\{(1, 6)\}) = \frac{1}{36} 6) Passo 1 For P(I_E=1): For the sample space, (HHHHH). P(I_E=1) = P(HHHHH) = \frac{1}{2^5} = \frac{1}{2×2×2×2×2} = \frac{1}{32} Resultado P(I_E=1) = \frac{1}{32} 10) Passo 1 Let X be the random variable for the number of sixes, X = 0, 1, 2, 3. P(X = 0) + P(X = 1) Let N be the event of getting a six on a single throw. P(N) = \frac{1}{6} Passo 2 Let N^c be the event of not getting a six. P(N^c) = \frac{5}{6} P(X = 0) = P(N^cN^cN^c) = P(N^c)P(N^c)P(N^c) = \frac{5}{6}×\frac{5}{6}×\frac{5}{6} = \frac{125}{216} For P(X = 1): P(X = 1) = P(N^cN^cN) + P(N^cNN^c) + P(NN^cN^c) = P(N^c)P(N^c)P(N) + P(N^c)P(N)P(N^c) + P(N)P(N^c)P(N^c) = \frac{5}{6}×\frac{5}{6}×\frac{1}{6} + \frac{5}{6}×\frac{1}{6}×\frac{5}{6} + \frac{1}{6}×\frac{5}{6}×\frac{5}{6} = \frac{75}{216} Passo 3 For the required probability: P(X = 0) + P(X = 1) = \frac{125}{216} + \frac{75}{216} = \frac{25}{27} Resultado The required probability: \frac{25}{27} 11) Passo 1 Let P be the probability of getting a white ball. P = \frac{1}{2} and, 1-P = \frac{1}{2} Passo 2 Let N the number of white balls that appeared. P(N = 2) = \frac{4}{2} (P^2)(1-P)^{4-2} = \frac{4!}{(2!)(2!)} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^2 = 6×\frac{1}{6} = \frac{3}{8} Resultado P(N = 2) = \frac{3}{8} Passo 1 Since the problem is discrete, we will use recessive relation: \Rightarrow \frac{P(X=k)}{P(X=k-1)} = \frac{\binom{n}{k}p^{k}(1-p)^{n-k}}{\binom{n}{k-1}p^{k-1}(1-p)^{n-k+1}} \Rightarrow \frac{P(X=k)}{P(X=k-1)} = \frac{\frac{n!}{k!(n-k)!}p^{k}(1-p)^{n-k}}{\frac{n!}{(k-1)!(n-(k-1))!}p^{k-1}(1-p)^{n-k+1}} \Rightarrow \frac{P(X=k)}{P(X=k-1)} = \frac{(n-k+1)p}{k(1-p)} \Rightarrow \frac{P(X=k)}{P(X=k-1)} \ge 1 \Rightarrow \frac{(n-k+1)p}{k(1-p)} \ge 1 \Rightarrow np-kp+p \ge k \Rightarrow np+p \ge k \Rightarrow p(n+1) \ge k Passo 2 2 de 4 (a) For (n+1)p is an integer; Let m be an integer. \frac{p(k)}{p(k-1)} = 1 + \frac{m-k}{k(1-p)} That concludes; \frac{p(k)}{p(k-1)} For > 1 k = 1, 2, ..., m-1 For = 1 k = m For < 1 k = m+1, m+2, ..., n Therefore if we say that (n+1)p is an integer, it shows that P(X=k) increases until k = m-1 and p(X=k-1) = p(X=k) and then after that is starts to decrease. Passo 3 3 de 4 (b) For (n+1) is not an integer; Let m be an integer and y a fractional, 0 < y < 1. \frac{p(k)}{p(k-1)} = 1 + \frac{(m-y)-k}{k(1-p)} That concludes; \frac{p(k)}{p(k-1)} For > 1 k = 0, 1, 2, ..., m For < 1 k = m+1, m+2, ..., n Therefore if we say that (n+1)p is not an integer, it shows that P(X=k) increases until k = m and then after that is starts to decrease. 16) Passo 1 Let p the probability that a person will not show up. p=0.05 and, 1-p = 1-0.005 p=0.95 Passo 2 2 de 3 So to find the probability that there will be a seat left for every passenger who will show up, atleast two persons should not show up. If we let X be the binomial random variable for the number of persons who will not show up, the required probability will be; P(X \ge 2) = 1 - P(X < 2) = 1 - P(X = 0) + P(X = 1) = 1 - \binom{52}{0}(0.05)^{0}(1-0.05)^{52} + \binom{52}{1}(0.05)^{1}(1-0.005)^{52-1} = 1 - 1 \times 1 \times (0.95)^{52} + 52 \times 0.05 \times (0.95)^{51} = 1 - 0.259497 = \boxed{0.7405} Resultado P = 0.7405 20) Passo 1 Using multinomial probability distribution: Denote P_1 = 0.5, P_2 = 0.2, P_3 = 0.3. n=5; x_1=1, x_2=2, x_3=2 \Rightarrow P(x_1=1, x_2=2, x_3=2) = \frac{5!}{(1!)(2!)(2!)} [(0.5)^{1}(0.2)^{2}(0.3)^{2}] = \frac{27}{500} Resultado P = \frac{27}{500} 21) Passo 1 1 de 4 The probability that a person will buy a television P_1 = 0.7, and for customers who purchase nothing P_2 = 0.3 and for the number of customer, n = 5. For the probability that the owner will sell three or more televisions: P(X ≥ 3) = P(3 ≥ X ≥ 5) = P(X = 3, 4, 5) ⟹ P(X = 3) + P(X = 4) + P(X = 5) Passo 2 2 de 4 For P(X = 3): P(X = 3) = \(\frac{5!}{(3!)(2!)}\) (0.7)^3 (0.3)^2 = 0.3087 For P(X = 4): P(X = 4) = \(\frac{5!}{(4!)(1!)}\) (0.7)^4 (0.3)^1 = 0.36015 Passo 3 3 de 4 For P(X = 5): P(X = 5) = \(\frac{5!}{(5!)(0!)}\) (0.7)^5 (0.3)^0 = 0.16807 P = 0.30870 + 0.36015 + 0.16807 = 0.83692 Resultado 4 de 4 P = 0.83692 31) Passo 1 1 de 5 (a) For n = 8, p = 0.1: λ = np = 8 × 0.1 = 0.8 The approximate probability using Poisson PMF: P(X = 2) = e^{-λ} \(\frac{λ^2}{2!}\) = e^{-0.8} \(\frac{(0.8)^2}{2!}\) = 0.14379 The exact probability using binomial PMF: P(X = 2) = \({n\choose x}\) (p)^x(1 − p)^{n−x} = \(\frac{8}{2}\) (0.1)^2 (1 − 0.1)^{8−2} = 0.1488 Passo 2 2 de 5 (b) For n = 10, p = 0.95: λ = np = 10 × 0.95 = 9.5 The approximate probability using Poisson PMF: P(X = 9) = e^{-λ} \(\frac{λ^9}{9!}\) = e^{-9.5} \(\frac{(9.5)^9}{9!}\) = 0.1300 The exact probability using binomial PMF: P(X = 9) = \({n\choose x}\) (p)^x(1 − p)^{n−x} = \(\frac{10}{9}\) (0.95)^9 (1 − 0.95)^{10−9} = 0.3151 Passo 3 3 de 5 (c) For n = 10, p = 0.1: λ = np = 10 × 0.1 = 1 The approximate probability using Poisson PMF: P(X = 0) = e^{-λ} \(\frac{λ^0}{0!}\) = e^{-1} \(\frac{(1)^0}{0!}\) = 0.3679 The exact probability using binomial PMF: P(X = 0) = \({n\choose x}\) (p)^x (1 − p)^{n−x} = \(\frac{10}{0}\) (0.1)^0 (1 - 0.1)^{10−0} = 0.3487 Passo 4 4 de 5 (d) For n = 9, p = 0.2: λ = np = 9 × 0.2 = 1.8 The approximate probability using Poisson PMF: P(X = 4) = e^{-λ} \(\frac{λ^4}{4!}\) = e^{-1.8} \(\frac{(1.8)^4}{4!}\) = 0.0723 The exact probability using binomial PMF: P(X = 4) = \({n\choose x}\) (p)^x (1 − p)^{n−x} = \(\frac{9}{4}\) (0.2)^4 (1 - 0.2)^{9−4} = 0.3457 32) Passo 1 1 de 5 For the desired probability, we can use Poisson to Binomial: \lambda = np = (50)\left( \frac{1}{100} \right) = 0.5 \implies P(X = x) = \frac{(e^{-0.5})(0.5)^{x}}{x!} , x = 1, 2, .... Passo 2 2 de 5 (a) For the probability of winning the prize at least once: P(X \ge 1) = (1 - P(X = 0)) = 1 - \frac{(e^{-0.5})(0.5)^{0}}{0!} = 1 - e^{-0.5} \approx [0.3935] Passo 3 3 de 5 (b) For the probability of winning the prize exactly once: P(X = 1) = (P)(X = 1) = \frac{(e^{-0.5})(0.5)^{1}}{1!} = 0.5e^{-0.5} \approx [0.3033] Passo 4 4 de 5 (c) For the probability of winning the prize at least twice: P(X \ge 2) = (1 - P(X = 0) - P(X = 1)) = 1 - \left( \frac{(e^{-0.5})(0.5)^{0}}{0!} \right) - \left( \frac{(e^{-0.5})(0.5)^{1}}{1!} \right) = 1 - e^{-0.5} - 0.5e^{0.5} \approx [0.0902] Resultado 5 de 5 (a) P(X \ge 1) = 0.3935 (b) P(X = 1) = 0.3033 (c) P(X \ge 2) = 0.0902 39) Passo 1 1 de 2 For E[X]: E[X] = \sum x[P(x)] = \left[ 1 \cdot \frac{1}{2} \right] + \left[ 2 \cdot \frac{1}{3} \right] + \left[ 24 \cdot \frac{1}{6} \right] = \left[ \frac{1}{2} \right] + \left[ \frac{2}{3} \right] + 4 = \left[ \frac{3 + 3 + 4 + 24}{6} \right] = \frac{31}{6} \approx [5.1667] Passo 2 2 de 4 For the expected number of games: 𝐸[𝑋]=∑𝑖=47𝑖[𝑃{𝑋=𝑖}]=4[𝑝4+(1−𝑝)4]+5[4𝑝3(1−𝑝)𝑝+4(1−𝑝)3𝑝(1−𝑝)]+6 [10𝑝3(1−𝑝)2𝑝+10(1−𝑝)3𝑝2(1−𝑝)]+7[20𝑝3(1−𝑝)3]E[X]=i=4∑7 i[P{X=i}]=4[p4+(1−p)4]+5[4p3(1−p)p+4(1−p)3p(1−p)]+6[10p3(1−p)2p+ 10(1−p)3p2(1−p)]+7[20p3(1−p)3] Passo 3 quando p = \frac{1}{2} = 4 \left[ \left( \frac{1}{2} \right)^{4} + \left( \frac{1}{2} \right)^{1} \right] + 5 \left[ \left( 4 \times \frac{1}{2} \right) \times \left( 1 - \frac{1}{2} \right) + 4 \times \left( \frac{1}{2} \right) \times \frac{1}{2} \times \left( 1 - \frac{1}{2} \right) \right] + 6 \left[ 10 \times \left( \frac{1}{2} \right)^{3} \times \left( 1 - \frac{1}{2} \right)^{2} \times \frac{1}{2} + 10 \times \left( 1 - \frac{1}{2} \right) \times \left( \frac{1}{2} \right)^{2} \times \left( 1 - \frac{1}{2} \right) \right] + 7 \left[ 20 \times \left( \frac{1}{2} \right)^{3} \times \left( 1 - \frac{1}{2} \right)^{3} \right] = \frac{93}{16} \approx [5.8125]