8.3 Capítulo 3

Solução. 3.4

  1. \(\Omega_{4} = \{ 1,2,3,4 \}\)
    \(\Omega_{6} = \{ 1,2,3,4,5,6 \}\)
    \(\Omega_{8} = \{ 1,2,3,4,5,6,7,8 \}\)
    \(\Omega_{10} = \{ 1,2,3,4,5,6,7,8,9,10 \}\)
    \(\Omega_{12} = \{ 1,2,3,4,5,6,7,8,9,10,11,12 \}\)
    \(\Omega_{20} = \{ 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 \}\)
    \(\Omega_{100} = \{ 10,20,30,40,50,60,70,80,90,100 \}\)

  2. \[ \Omega_{4,4} = \left\lbrace \begin{array}{cccccc} (1,1) & (1,2) & (1,3) & (1,4) \\ (2,1) & (2,2) & (2,3) & (2,4) \\ (3,1) & (3,2) & (3,3) & (3,4) \\ (4,1) & (4,2) & (4,3) & (4,4) \\ \end{array} \right\rbrace \]

  3. \[ \Omega_{4,6} = \left\lbrace \begin{array}{cccccc} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ \end{array} \right\rbrace \]

Solução. 3.5
a. \(Pr(C) = \frac{1}{308,000} \approx 0.0000032468\)
b. \(Pr(H) = \frac{1}{17,600} \approx 0.0000568182\)
c. \(Pr(M) = \frac{1}{10,600} \approx 0.0000943396\)
d. \(Pr(C \cup H \cup M) = \frac{1}{308,000} + \frac{1}{17,600} + \frac{1}{10,600} = \frac{5,041}{32,648,000} \approx 0.0001544046\)
e. \(8,000,000,000 \times \frac{1}{308,000} \approx 25,974\)
\(8,000,000,000 \times \frac{1}{17,600} \approx 454,545\)
\(8,000,000,000 \times \frac{1}{10,600} \approx 754,717\)
\(8,000,000,000 \times \frac{5,041}{32,648,000} \approx 1,235,236\)

Solução. 3.6
a. \(\frac{1}{2} + \frac{1}{3} + \frac{1}{9}= \frac{17}{18} < 1\)
b. Como o resto da divisão de 35 por 3 é 2, sobram 2 camelos não importa como seja feita a divisão. Note que \(18+12+4=34\).

Solução. 3.7

  • P4 (Complementar)
    Por (3.22) \[ Pr(\Omega) = 1 \] Pode-se escrever \(\Omega = A \cup A^c\), então \[ Pr(A \cup A^c) = 1 \] Como \(A \cap A^c = \emptyset\), por (3.23)
    \[ Pr(A \cup A^c) = Pr(A) + Pr(A^c) \] Desta forma, \[ Pr(A) + Pr(A^c) = 1 \therefore Pr(A) = 1 - Pr(A^c) \] \(\bigtriangleup\)

  • P5

  • P6
    \(A \cup B = A \cup (B-A)\) e \(A \cap (B-A) = \emptyset\). Logo, por (3.23)
    \[\begin{equation} Pr(A \cup B) = Pr(A) + Pr(B-A) \tag{1} \end{equation}\] Como \(A \subset B\), \(B = (B-A) \cup (A \cap B)\) e \((B-A) \cap (A \cap B) = \emptyset\). Logo, novamente por (3.23)
    \[\begin{equation} Pr(B) = Pr(B-A) + Pr(A \cap B) \tag{2} \end{equation}\] Combinando (1) e (2), \[ Pr(A \cup B) = Pr(A) + Pr(B) - Pr(A \cap B). \] \(\bigtriangleup\)

  • P7

  • P8

Solução. 3.11

Solução. 3.13

  1. \(R_D = \{ -5,-4,-3,-2,-1,0,1,2,3,4,5 \}\)

  2. \(R_P = \{ 1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,30,36 \}\)

  3. \(R_Q = \left\{ \frac{1}{6},\frac{1}{5},\frac{1}{4},\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},1,\frac{6}{5},\frac{5}{4},\frac{4}{3},\frac{3}{2},\frac{5}{3},2,\frac{5}{2},3,4,5,6 \right\}\)

Solução. 3.14

  2. \(R_Y = \{ -5,-4,-3,-2,-1,0,1,2,3,4,5 \}\)
    \(Pr(Y=-5) = Pr(Y=5) = \frac{1}{36}\)
    \(Pr(Y=-4) = Pr(Y=4) = \frac{2}{36}\)
    \(Pr(Y=-3) = Pr(Y=3) = \frac{3}{36}\)
    \(Pr(Y=-2) = Pr(Y=2) = \frac{4}{36}\)
    \(Pr(Y=-1) = Pr(Y=1) = \frac{5}{36}\)
    \(Pr(Y=0) = \frac{6}{36}\)
    De forma geral, \(Pr(Y=y) = \frac{6-|y|}{36}\).

Solução. 3.15

Solução. 3.18

  1. Sabe-se que \(\sum_{x=0}^{\infty} \frac{\lambda^x}{x!} = e^\lambda\). Assim,

\[\begin{align*} \sum_{x=0}^{\infty} \left[ p \frac{e^{-\lambda_1} \lambda_{1}^{x}}{x!} + (1-p) \frac{e^{-\lambda_2} \lambda_{2}^{x}}{x!} \right] =& p e^{-\lambda_1} \sum_{x=0}^{\infty} \frac{\lambda_{1}^{x}}{x!} + e^{-\lambda_2} \sum_{x=0}^{\infty} \frac{\lambda_{2}^{x}}{x!} - p e^{-\lambda_2} \sum_{x=0}^{\infty} \frac{\lambda_{2}^{x}}{x!} \\ =& p e^{-\lambda_1} e^{\lambda_1} + e^{-\lambda_2} e^{\lambda_{2}} - p e^{-\lambda_2} e^{\lambda_{2}} \\ =& p + 1 - p \\ =& 1 \end{align*}\] \(\bigtriangleup\)

Solução. 3.25

  1. \(F(X) = 1-e^{-2x}\)

  2. Pr(X<1) = F(1) = 1-e^{-2 } = 1-e^{-2}

  3. Pr(X>2) = 1-F(2) = 1-(1-e^{-2 }) = e^{-4} 

curve(dexp(x), 0, 4)