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# System prepended metadata

title: 機率HW6

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機率HW6
===

**All answers are verified by textbooks.**

## 1.
$C^6_3 \times (\frac{1}{6})^3 \times (\frac{5}{6})^3 = 0.002679184\times 20 =0.053583676$

### A:++0.05358367++
## 2.

$p = C^{10}_k(0.45)^k(0.55)^{10-k}$
$p= 252 \times 0.018452812 \times 0.050328438 = 0.234032704 \ when \ k=5$
$p = 210 \times 0.04100625 \times 0.027680641 = 0.238366647 \ when \ k=4$

### A:++k=4 , 0.238366647++

## 3.
$\Sigma^4_{i=1} C^8_{2i} (0.001)^{2i} (0.999)^{8-2i} = 0.000028$

### A:++0.000028++

## 4.
$\lambda  = 80 \times 0.025 = 2$
$1 - P(X=1) - P(X=0) = 1 - \frac{e^{-\lambda}\lambda^1}{1!} - \frac{e^{-\lambda}\lambda^0}{0!} = 1 - 3e^{-\lambda} = 0.594$

### A:++0.594++

## 5.
$\lambda =\frac{3}{10} \times 35 = 10.5$(One chapter)
$\frac{e^{-10.5} 10.5^{10}}{10!} = 0.124$
$0.124^2 = 0.0154$

### A:++0.0154++

## 6.
$P(X=1)=P(X=3)$
$\frac{e^{-\lambda}\lambda^1}{1!} = \frac{e^{-\lambda}\lambda^3}{3!}$
$6e^{-\lambda}\lambda^1 = e^{-\lambda}\lambda^3$
$\ln(6e^{-\lambda}\lambda^1) = \ln(e^{-\lambda}\lambda^3)$
$\ln 6 + \ln e^{-\lambda} + \ln \lambda^1 =\ln e^{-\lambda} + \ln \lambda^3$
$\ln 6  -\lambda + \ln \lambda^1 = -\lambda + \ln \lambda^3$
$\ln 6 + \ln \lambda^1 = \ln \lambda^3$
$\ln 6 = \ln\lambda^2$
$\lambda^2=6$
$\lambda=\sqrt{6}$

$P(X=5) = \frac{e^{-\sqrt{6}} \sqrt{6}^5}{5!} = 0.063$

### A:++0.063++

## 7.
$0.2 \times C^7_2 \times 0.2^2 \times 0.8^5 \simeq 0.055$

### A:++0.055++

## 8.
Need at least n bulbs,so n-1 bulbs are defective for sure,so answer is $p^{n-1}$.

### A: $p^{n-1}$

## 9.
$P(X=i) = 0.15 \times C^{i+9}_9 0.15^9 0.85^i$

## 10.
The probability of 4 tagged trout among the second 50 trout caught is
$P_n = \frac{C^{50}_4 C^{n-50}_{46}}{C^n_{50}}$

$\frac{P_n}{P_{n-1}} = \frac{(n-50)^2}{n(n-96)}$

If $P_n \ge P_{n-1}$ then $\frac{P_n}{P_{n-1}} \ge 1$ then $(n-50)^2 \ge n(n-96),$,so $n \le 625$.
If $P_n \le P_{n-1}$ then $\frac{P_n}{P_{n-1}} \le 1$ then $(n-50)^2 \le n(n-96),$,so $n \ge 625$.
$625 \le n \le 625$,so $n=625$

A:++625++



