# 機率HW5
## 110590064 劉韶軒
## Problem1
> Two fair dice are rolled and the absolute value of the difference of the outcomes is denoted by X. What are the possible values of X, and the probabilities associated with them?
## Ans

| 絕對值差 | 情況 | 機率 |
|:-------- |:----------------------------------------------------------- | --- |
| 0 | {1,1},{2,2},{3,3},{4,4},{5,5},{6,6} | $\frac{6}{36}$ |
| 1 | {1,2},{2,3},{3,4},{4,5},{5,6} {6,5},{5,4},{4,3},{3,2},{2,1} | $\frac{10}{36}$ |
| 2 | {1,3},{2,4},{3,5},{4,6},{6,4},{5,3},{4,2},{3,1} | $\frac{8}{36}$ |
| 3 | {1,4},{2,5},{3,6},{6,3},{5,2},{4,1} | $\frac{6}{36}$ |
| 4 | {1,5},{2,6},{6,2},{5,1} | $\frac{4}{36}$ |
| 5 | {1,6},{6,1} | $\frac{2}{36}$ |

$F(X)=2(6-X)$
$P(X)=\frac{2(6-X)}{36}$

## Problem 2
> F, the distribution function of a random variable X, is given by
Then, calculate the following quantities: P(X < 1), P(X = 1), P(0 ≤ X < 1), P(X > 1/2), P(X = 3/2), and P(1 < X ≤ 6).

$P(X < 1)=\frac{1}{2}$
$P(0 ≤ X < 1)=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$
$P(X > \frac{1}{2})=1-\frac{1}{2}=\frac{1}{2}$
$P(X = \frac{3}{2})=1$
$P(1 < X ≤ 6)=1-\frac{1}{2}=\frac{1}{2}$

## Problem 3
> Assume that 12% of people have traveled internationally. A company wants to hire a new employee who has traveled internationally. How many applicants do they need to interview to have a 60% chance that at least one of the applicants has traveled internationally??

## Ans
P(面試了n個人都沒出過國的機率)=$0.88^n$
$1-P(N)≧0.6=P(N)≤0.4$
$P(N)≤0.4$
$n*log(0.88)≤log(0.4)$
$n≤\frac{log(0.4)}{log(0.88)}$
$n≤7.167$
$n得7$
## Problem 4
> In the experiment of rolling a balanced die twice, let X be the minimum of the two numbers obtained. Determine the probability mass function and the distribution function of X
## Ans

| min | 情況 | 機率 |
|:--- |:----------------------------------------------------------------- | --------------- |
| 1 | {1,1},{1,2},{1,3},{1,4},{1,5},{1,6} {6,1},{5,1},{4,1},{3,1},{2,1} | $\frac{11}{36}$ |
| 2 | {2,3},{2,4},{2,5},{2,6},{6,2},{5,2},{4,2},{3,2},{2,2} | $\frac{9}{36}$ |
| 3 | {3,4},{3,5},{3,6},{6,3},{5,3},{4,3},{3,3} | $\frac{7}{36}$ |
| 4 | {4,5},{4,6},{6,4},{5,4},{4,4} | $\frac{5}{36}$ |
| 5 | {5,6},{6,5},{5,5} | $\frac{3}{36}$ |
| 6 | {6,6} | $\frac{1}{36}$ |

| mass function | 情況 | 機率 |
|:------------- |:----------------------------------------------------------------- | --------------- |
| <=1 | {1,1},{1,2},{1,3},{1,4},{1,5},{1,6} {6,1},{5,1},{4,1},{3,1},{2,1} | $\frac{11}{36}$ |
| <=2 | {2,3},{2,4},{2,5},{2,6},{6,2},{5,2},{4,2},{3,2},{2,2}+前面 | $\frac{20}{36}$ |
| <=3 | {3,4},{3,5},{3,6},{6,3},{5,3},{4,3},{3,3}+前面 | $\frac{27}{36}$ |
| <=4 | {4,5},{4,6},{6,4},{5,4},{4,4}+前面 | $\frac{32}{36}$ |
| <=5 | {5,6},{6,5},{5,5}+前面 | $\frac{35}{36}$ |
| <=6 | {6,6}+前面 | $\frac{36}{36}$ |
## Problem 5
> In a lottery every week, 2,000,000 tickets are sold for $1 apiece. If 4000 of these tickets pay off $30 each, 500 pay off $800 each, one ticket pays off $1,200,000, What is the expected value of the winning amount for a player with a single ticket?
## Ans
$\frac{4000}{2,000,000}*30+\frac{500}{2,000,000}*800+\frac{1}{2,000,000}*1200000=-\frac{7}{50}$

## Problem 6
> If X is a random number selected from the first 10 positive integers, what is E\[X(11-X)\]?

## Ans
$P(X=k)=\frac{1}{10},\text{for k} \in {1,2,3..10}$
$E(X(11-X))=E(11X)-E(X^2)$
$E(X)=\sum_{X\in A}X * p(X)$
$E(X)=\sum_{1}^{10}X * 0.1=5.5$
$E(X^2)=\sum_{1}^{10}X * 0.1=385$
$E(X(11-X))=55-385=-330*
## Problem 7
> Mr. Jones is about to purchase a business. There are two businesses available. The first has a daily expected profit of $150 with standard deviation $30, and the second has a daily expected profit of $150 with standard deviation $55. If Mr. Jones is interested in a business with a steady income, which should he choose?

## Ans
應該選擇公司1，在兩者獲利情況一致下，我們比較標準差，如果標準差較大，代表波動較大，他想要穩定，所以使用標準差越小越好

## Problem 8
> Find the variance and the standard deviation of a random variable X with distribution function

## Ans
$Var(X) = E(X^2)– (E(X))^2$
$x={-3,0,6}$
$P(-3)=\frac{3}{8}-0$
$P(0)=\frac{3}{4}-\frac{3}{8}=\frac{3}{8}$
$P(0)=1-\frac{3}{4}=\frac{1}{4}$
$E(X)=-3*\frac{3}{8}+6*\frac{1}{4}=\frac{3}{8}$
$E(X^2)=9*\frac{3}{8}+36*\frac{1}{4}=\frac{99}{8}=12.375$
$var(X)=12.375$
$Std(X)=\sqrt{12.375}=3.51781181987$

## Problem 9
> Suppose that X is a discrete random variable with E\[X\]=1 and E\[X(X-2)\]=3. Find Var(-3X+7)=?
## Ans
$E[X]=1$
$E[X(X-2)]=3$
$E[X^2]+E[-2X]=3$
$E[X^2]-2E[X]=3$
$E[X^2]=5$
$Var(-3X+7)$
$Var(aX+b)=a^2Var(X)$
$Var(-3X+7) = (-3)^2Var(X) = 9Var(X)$
$Var(X) = E(X^2) - (E(X))^2$
$E(X^2) = E[X(X-2)] + 2E(X) = 3 + 2 = 5$
$Var(X) = E(X^2) - (E(X))^2 = 5 - 1^2 = 4$。$
$Var(-3X+7) = 9Var(X) = 9*4=36$。