자연계열 (오후1) 1번

1.숫자 3이 나오는 횟수가 2의 배수일 확률 $\displaystyle ={}_n\mathrm C_0\left(\frac16\right)^0\left(\frac56\right)^n+{}_n\mathrm C_2\left(\frac16\right)^2\left(\frac56\right)^{n-2}+{}_n\mathrm C_4\left(\frac16\right)^4\left(\frac56\right)^{n-4}+\cdots\quad-(\text{★})$$\displaystyle \begin{aligned}\left(\frac16+\frac56\right)^n&={}_n\mathrm C_0\left(\frac16\right)^0\left(\frac56\right)^n+{}_n\mathrm C_1\left(\frac16\right)^1\left(\frac56\right)^{n-1}+\cdots\\+\left(\left(-\frac16\right)+\frac56\right)^n&={}_n\mathrm C_0\left(-\frac16\right)^0\left(\frac56\right)^n+{}_n\mathrm C_1\left(-\frac16\right)^1\left(\frac56\right)^{n-1}+\cdots\\\hline&=2\times\text{★}\end{aligned}$

$\displaystyle \therefore\text{★}=\frac12\left(1+\left(\frac23\right)^n\right)$㈜ 0을 2의 배수로 생각하지 않는 학생이 있을 수 있다. 이 경우 답은, ★에서 0회일 확률을 제하여,$\displaystyle \frac12\left(1+\left(\frac23\right)^n\right)-\left(\frac56\right)^n$

2.기댓값 $\displaystyle =\sum_{k=3}^n100\times k(k-1)(k-2)\times{}_n\mathrm C_k\times\left(\frac16\right)^k\left(\frac56\right)^{n-k}=\frac{100}{6^n}\sum_{k=3}^nk(k-1)(k-2)\times{}_n\mathrm C_k\times5^{n-k}$$\displaystyle k(k-1)(k-2)\times{}_n\mathrm C_k=k(k-1)(k-2)\times\frac{n!}{k!(n-k)!}=\frac{(n-3)!}{(k-3)!((n-3)-(k-3))!}\times n(n-1)(n-2)$$\displaystyle \begin{aligned}\therefore\text{기댓값}&=\frac{100\times n(n-1)(n-2)}{6^n}\sum_{k=3}^n{}_{n-3}\mathrm C_{k-3}\times5^{n-k}\\&=\frac{100\times n(n-1)(n-2)}{6^n}\sum_{k=0}^{n-3}{}_{n-3}\mathrm C_k\times5^{n-(k+3)}\quad\cdots\cdots(5^{n-(k+3)}=5^{(n-3)-k})\end{aligned}$

$\displaystyle =\frac{100\times n(n-1)(n-2)}{6^n}(1+5)^{n-3}=\frac{100\times n(n-1)(n-2)}{6^3}=\frac{25}{54}n(n-1)(n-2)$ (원)

3.$\displaystyle \begin{aligned}\text{A가 받을 금액의 기댓값}&=3\times0\times{}_n\mathrm C_0\times\left(\frac16\right)^0\left(\frac56\right)^n+3\times2\times{}_n\mathrm C_2\times\left(\frac16\right)^2\left(\frac56\right)^{n-2}+\cdots\\-\text{) B가 받을 금액의 기댓값}&=3\times1\times{}_n\mathrm C_1\times\left(\frac16\right)^1\left(\frac56\right)^{n-1}+3\times3\times{}_n\mathrm C_3\times\left(\frac16\right)^3\left(\frac56\right)^{n-3}+\cdots\end{aligned}$$\displaystyle \begin{aligned}&=\sum_{k=0}^n3k\times{}_n\mathrm C_k\times\left(-\frac16\right)^k\left(\frac56\right)^{n-k}=\frac3{6^n}\sum_{k=0}^nk\times{}_n\mathrm C_k\times(-1)^k5^{n-k}=\frac3{6^n}\sum_{k=1}^nk\times{}_n\mathrm C_k\times(-1)^k5^{n-k},\quad(k\times{}_n\mathrm C_k=n\times{}_{n-1}\mathrm C_{k-1}\text{이므로})\\&=\frac{3n}{6^n}\sum_{k=1}^n{}_{n-1}\mathrm C_{k-1}\times(-1)^k\times5^{n-k}=\frac{3n}{6^n}\sum_{k=0}^{n-1}{}_{n-1}\mathrm C_k\times(-1)^{k+1}\times5^{(n-1)-k}=\frac{-3n}{6^n}\sum_{k=0}^{n-1}{}_{n-1}\mathrm C_k\times(-1)^k\times5^{(n-1)-k}\\&=\frac{-3n}{6^n}(-1+5)^{n-1}=\frac{-3n}{6^n}4^{n-1}=\frac{-n}2\left(\frac23\right)^{n-1}\end{aligned}$

∴ B쪽이 더 크다. 차이 $\displaystyle =\frac n2\left(\frac23\right)^{n-1}$ (원)