# 309553032_林育駿_hw03

###### tags: `Biostatistics`

## 1.
### (a)
$P(B|A) = \frac{P(A|B)\times P(B)}{P(A)}=\frac{\frac{P(A\cap B)}{P(B)}\times P(B)}{P(A)}=\frac{0.082}{0.138}=59.42\%$
### (b)
$P(A|B)=\frac{P(B|A)\times P(A)}{P(B)}=\frac{\frac{P(A\cap B)}{P(A)}\times P(A)}{P(B)}=\frac{P(A\cap B)}{P(B)}=\frac{0.082}{0.261}=31.42\%$
### (c\)
No.
One definition of independence between two events A and B is that
$$P(A|B)=P(A)=P(A|U),\ U:universe$$
In this case, $P(A)=0.138\neq 0.314=P(A|B)$, so events A and B are not independent.

## 2.
### (a)
因資料來源不同，不同 column 不可相加。
### (b)
$P(E|D)=\frac{273}{989}=27.60\%$

$P(E|D')=\frac{2641}{9901}=26.67\%$

the ratio is $\frac{27.60}{26.67}=1.03$

E與D呈現正相關，因為E在D中的比例比在D'中更高。

## 3.
### (a)
D: Complication
E: Fragmin
$RR = \frac{P(D|E)}{P(D|E^c)}=\frac{42/1518}{73/1473}=\frac{42\times 1473}{1518\times 73}=55.83\%$
### (b)
odds for exposed group: $\frac{P(D|E)}{P(D^c|E)}=\frac{42/1518}{1476/1518}=\frac{42}{1476}=0.02845$

odds for non-exposed group: $\frac{P(D|E^c)}{P(D^c|E^c)}=\frac{73}{1473}=0.04955$

odds ratio: $\frac{0.02845}{0.04955}=57.42\%$, which is far from 1, it means that influences the disease status.

## 4.
### (a)
$1-(0.05+0.1+0.6+0.15)=0.1$
### (b)
$P(X\leq 8)=0.75$
The probability of $X\leq 8=0.75$
### (c\)
$P(X\lt 8)=0.15$
### (d)
$P(X\geq 7)=0.95$
### (e)
$P(X\gt 7)=0.85$

## 5.
goal is to solve $P(C|D)$.

$P(C|D)=\frac{P(D|C)\times P(C)}{P(D)}$, we have $P(D|C), P(C)$, now we have to find $P(D)$

$\because U=A\cup B\cup C$

$\therefore P(D)=P(D\cap A)+P(D\cap B)+P(D\cap C)$

$=P(D|A)\times P(A)+P(D|B)\times P(B)+P(D|C)\times P(C)$

$=0.025\times 0.40+0.015\times 0.30
+0.012\times 0.30$

$=0.0181$

$P(C|D)=\frac{P(D|C)\times P(C)}{P(D)}=\frac{0.012\times 0.30}{0.0181}=19.89\%$