자연계열 3번

3-1.$\displaystyle a\ge b$일 때, $\displaystyle a=n$일 확률은 조건부확률$\displaystyle P_n=\mathrm{P}(a=n\mid a\ge b)=\frac{\mathrm{P}(a=n\cap a\ge b)}{\mathrm{P}(a\ge b)}$ $\displaystyle (n=0,1,2,3,4)$이다.그런데, $\displaystyle \mathrm{P}(a\ge b)=\sum_{n=0}^4\mathrm{P}(a=n\cap a\ge b)$이므로 각 항을 구하여 보면,

1) $\displaystyle \mathrm{P}(a=0\cap a\ge b)=\mathrm{P}(a=0)\cdot\mathrm{P}(a\ge b\mid a=0)$으로$\displaystyle \mathrm{P}(a=0)=\left(\frac12\right)^4=\frac1{16}$이고,$\displaystyle a=0$일 때 $\displaystyle a\ge b$이면, $\displaystyle b=0$뿐 이므로$\displaystyle \mathrm{P}(a\ge b\mid a=0)=\mathrm{P}(b=0)=\frac14$$\displaystyle \therefore\mathrm{P}(a=0\cap a\ge b)=\frac1{64}$

2) $\displaystyle \mathrm{P}(a=1\cap a\ge b)=\mathrm{P}(a=1)\cdot\mathrm{P}(a\ge b\mid a=1)$으로$\displaystyle \mathrm{P}(a=1)={}_4\mathrm{C}_1\left(\frac12\right)^4=\frac4{16}$이고,$\displaystyle a=1$일 때 $\displaystyle a\ge b$이면, $\displaystyle b=0$ 또는 $\displaystyle 1$이므로$\displaystyle \mathrm{P}(a\ge b\mid a=1)=\frac34$$\displaystyle \therefore\mathrm{P}(a=1\cap a\ge b)=\frac{12}{64}$

3) $\displaystyle \mathrm{P}(a=2\cap a\ge b)=\mathrm{P}(a=2)\cdot\mathrm{P}(a\ge b\mid a=2)$으로$\displaystyle \mathrm{P}(a=2)={}_4\mathrm{C}_2\left(\frac12\right)^4=\frac6{16}$이고,$\displaystyle a=2$일 때 $\displaystyle a\ge b$이면, $\displaystyle b=0$ 또는 $\displaystyle 1$ 또는 $\displaystyle 2$이므로$\displaystyle \mathrm{P}(a\ge b\mid a=2)=1$$\displaystyle \therefore\mathrm{P}(a=2\cap a\ge b)=\frac6{16}=\frac{24}{64}$

4) $\displaystyle \mathrm{P}(a=3\cap a\ge b)=\mathrm{P}(a=3)\cdot\mathrm{P}(a\ge b\mid a=3)$으로$\displaystyle \begin{aligned}&=\mathrm{P}(a=3)\cdot1\\&={}_4\mathrm{C}_3\left(\frac12\right)^4=\frac4{16}=\frac{16}{64}\end{aligned}$$\displaystyle \therefore\mathrm{P}(a=3\cap a\ge b)=\frac4{16}=\frac{16}{64}$

5) $\displaystyle \mathrm{P}(a=4\cap a\ge b)=\mathrm{P}(a=4)\cdot1$으로$\displaystyle ={}_4\mathrm{C}_4\left(\frac12\right)^4=\frac1{16}=\frac4{64}$$\displaystyle \therefore\mathrm{P}(a=4\cap a\ge b)=\frac1{16}=\frac4{64}$따라서 $\displaystyle \mathrm{P}(a\ge b)=\frac1{64}+\frac{12}{64}+\frac{24}{64}+\frac{16}{64}+\frac4{64}=\frac{57}{64}$

그러므로 $\displaystyle P_n=\mathrm{P}(a=n\mid a\ge b)=\frac{\mathrm{P}(a=n\cap a\ge b)}{\mathrm{P}(a\ge b)}$ $\displaystyle (n=0,1,2,3,4)$은 다음과 같다.

$\displaystyle P_0=\frac{1/64}{57/64}=\frac1{57},\quad P_1=\frac{12/64}{57/64}=\frac{12}{57}=\frac4{19}$$\displaystyle P_2=\frac{24/64}{57/64}=\frac{24}{57}=\frac8{19},\quad P_3=\frac{16/64}{57/64}=\frac{16}{57}$$\displaystyle P_4=\frac{4/64}{57/64}=\frac4{57}$

3-2.$\displaystyle a\ge b$일 때, $\displaystyle a\ge c$일 확률은 조건부확률$\displaystyle \mathrm{P}(a\ge c\mid a\ge b)=\frac{\mathrm{P}(a\ge c\cap a\ge b)}{\mathrm{P}(a\ge b)}$이다.그런데, $\displaystyle \mathrm{P}(a\ge c\cap a\ge b)=\sum_{n=0}^4\mathrm{P}(a=n)\mathrm{P}(a\ge c\cap a\ge b\mid a=n)$이므로$\displaystyle \begin{aligned}\mathrm{P}(a\ge c\mid a\ge b)&=\sum_{n=0}^4\frac{\mathrm{P}(a=n)\mathrm{P}(a\ge c\cap a\ge b\mid a=n)}{\mathrm{P}(a\ge b)}\\&=\sum_{n=0}^4\frac{\mathrm{P}(a=n)\mathrm{P}(a\ge b\mid a=n)\mathrm{P}(a\ge c\mid a=n)}{\mathrm{P}(a\ge b)}=\sum_{n=0}^4\mathrm{P}(a=n\mid a\ge b)\mathrm{P}(a\ge c\mid a=n).\end{aligned}$문항 3-1에서 구한 $\displaystyle P_n=\mathrm{P}(a=n\mid a\ge b)$을 활용하면,

$\displaystyle P_0=\frac{1/64}{57/64}=\frac1{57},\quad P_1=\frac{12/64}{57/64}=\frac{12}{57}=\frac4{19}$$\displaystyle P_2=\frac{24/64}{57/64}=\frac{24}{57}=\frac8{19},\quad P_3=\frac{16/64}{57/64}=\frac{16}{57}$$\displaystyle P_4=\frac{4/64}{57/64}=\frac4{57}$

$\displaystyle a\ge b$일 때,$\displaystyle a=0$일 확률이 $\displaystyle \frac1{57}$,$\displaystyle a=1$일 확률이 $\displaystyle \frac{12}{57}$,$\displaystyle a=2$일 확률이 $\displaystyle \frac{24}{57}$,$\displaystyle a=3$일 확률이 $\displaystyle \frac{16}{57}$,$\displaystyle a=4$일 확률이 $\displaystyle \frac4{57}$이다.

$\displaystyle a\ge b$일 때, $\displaystyle a\ge c$일 확률은$\displaystyle \begin{aligned}\mathrm{P}(a\ge c\mid a\ge b)&=\mathrm{P}(a=0\mid a\ge b)\mathrm{P}(a\ge c\mid a=0)+\mathrm{P}(a=1\mid a\ge b)\mathrm{P}(a\ge c\mid a=1)\\&\quad+\mathrm{P}(a=2\mid a\ge b)\mathrm{P}(a\ge c\mid a=2)+\mathrm{P}(a=3\mid a\ge b)\mathrm{P}(a\ge c\mid a=3)\\&\quad+\mathrm{P}(a=4\mid a\ge b)\mathrm{P}(a\ge c\mid a=4)\end{aligned}$이다.$\displaystyle \mathrm{P}(a\ge c\mid a=0)=\mathrm{P}(0\ge c)=\mathrm{P}(0=c)=\frac14$$\displaystyle \mathrm{P}(a\ge c\mid a=1)=\mathrm{P}(1\ge c)=\frac34$$\displaystyle \mathrm{P}(a\ge c\mid a=2)=\mathrm{P}(2\ge c)=1$$\displaystyle \mathrm{P}(a\ge c\mid a=3)=P(a\ge c\mid a=4)=1$그러므로, $\displaystyle \begin{aligned}\mathrm{P}(a\ge c\mid a\ge b)&=\frac1{57}\times\frac14+\frac{12}{57}\times\frac34+\frac{24}{57}\times1+\frac{16}{57}\times1+\frac4{57}\times1\\&=\frac1{228}+\frac{36}{228}+\frac{176}{228}\\&=\frac{213}{228}=\frac{71}{76}\end{aligned}$