# 网传电子系概率论期末解答

Contents

2023-01-03更新

https://www.zhihu.com/question/419625606

## 1.

1. 这个cycle长度是1
2. 这个cycle长度是2
3. 这个 cycle长度是k
4. 这个cycle长度是N-1

\sum \frac{1}{N} \frac{1}{b_i} \frac{1}{b_{i+1}}\dots... = \sum_{1<i_1<i_2\dots<i_{N-k}} \frac{1}{N!} i_1i_2\dots i_{N-k}

\frac{1}{N!} (\sum_{1<i<N} i + \sum_{1<i_1<i_2<N}i_1i_2 +\dots + \sum_{1<i_1<i_2\dots < i_{N-2}})

N! = 2*(1+2)*(1+3)*\dots *(1+N-1)

\frac{N!}{2} = (1+2)*(1+3)*\dots *(1+N-1)

## 2.

n^2 - (4b+1)n + 2b^2+2b = 0

## 3.

F(s) = Prob\{S \le s\} = Prob\{ \frac{1}{2} r^2 |sin(\theta)| \le s \}

F(s) =Prob\{|sin(\theta)| \le \frac{2s}{r^2}\}

F(s) = Prob\{ 0 \le sin(\theta) \le \frac{2s}{r^2} \} + Prob\{ -\frac{2s}{r^2} \le sin(\theta) \le 0 \}

## 4.

• 从甲中取的俩球分别是A, B
• 事件I: 事后从乙中取了一个球是白色

P(A=B | I ) = \frac{ P(w,w,I) + P(b,b,I) }{P(w,w,I) + P(b,b,I) + P(w,b,I) + P(b,w,I)} = \frac{ P(w,w)P(I|w,w) + P(b,b)P(I|b,b) }{ P(w,w)P(I|w,w) + P(b,b)P(I|b,b) + P(w,b)P(I|w,b) + P(b,w)P(I|b,w) } = \frac{11}{41}

## 5.

X = \min \{ l, 1-l \} , 其中l \sim U(0, 1)

F(x) = Prob\{X \le x\} = 1 - Prob\{X > x\} = 1 - Prob\{ \min \{l,1-l\} > x \}

F(x) = 1 - Prob\{ l > x \& 1-l > x \} = 1 - Prob\{ x < l < 1-x \} = 1 - \int_{x}^{1-x}dx=2x, with\ x \in [0, \frac{1}{2}]

F(Y) = Prob\{Y\le y\} = Prob\{tan(X) \le y \} = Prob\{X \le arctan(y) \} = 2arctan(y)

f_y(y) = \frac{dF(Y)}{dy} = \frac{2}{1+y^2}

## 6.

E(Z) = \int_{-\infty}^{+\infty} cos(t) \frac{1}{2\sqrt{\pi}}e^{-\frac{t^2}{4}} dt

E(z) = \int_{-\infty}^{+\infty} (cos(t) + jsin(t)) \frac{1}{2\sqrt{\pi}}e^{-\frac{t^2}{4}} dt =\int_{-\infty}^{+\infty} e^{jt} \frac{1}{2\sqrt{\pi}}e^{-\frac{t^2}{4}} dt

\mathcal{F}(e^{jt} \frac{1}{2\sqrt{2\pi}}e^{-\frac{t^2}{4}} ) = \frac{1}{2\pi} 2\pi \delta(\omega-1) * e^{-\omega^2} = e^{-(\omega-1)^2}

## 7.

f(x|d) = \frac{f(x) f(d|x)}{f(d)}

E(X^2|D=d) = \int_{0}^{d} x^2 dx = \frac{1}{3}d^2

## 8.

E[N] = \sum_{N\ge 0}\frac{1}{N!} = e

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