115年台北科技大學電子碩士班-機率詳解

國立臺北科技大學115學年度碩士班招生考試

系所組別:電子工程 科目:機率

解答:$$\textbf{(一) }{P(A\cap C)\over P(C)} = \bbox[red, 2pt]{P(A\mid C)} \\ \textbf{(二) }A,B獨立 \Rightarrow P(A\cap B)= P(A)P(B)= 0.1\times 0.15=0.015\\ \qquad \Rightarrow \bbox[red, 2pt]{增加的問題:「您是否同時參加了 A 社團與 B 社團?」} \\ \qquad \bbox[red, 2pt]{\text{所需的比例:必須有 1.5% 的學生回答「是」（或者 98.5% 的學生回答「否」）}} \\ \textbf{(三) }\cases{P(A')=1-P(A)=0.9\\ P(B')=1-P(B)-0.85} \Rightarrow P(A'\cap B')= P(A')\times P(B')= \bbox[red, 2pt]{0.765}$$

解答:$$\textbf{(一) } E[X] =(-3)\times {1\over 2}+3\times {1\over 3}+6\times {1\over 6} =\bbox[red, 2pt]{1\over 2} \\ \textbf{(二) } E[X^2]=(-3)^2\times {1\over 2}+ 3^2\times {1\over 3}+6^2\times {1\over 6} ={27\over 2} \\\qquad \Rightarrow Var(X)=E[X^2]-(E[X])^2={27\over 2}-{1\over 4} =\bbox[red, 2pt]{53\over 4} \\ \textbf{(三)} E[(X-1)^2] =E[X^2-2X+1] =E[X^2]-2E[X]+E[1] ={27\over 2}-2\cdot {1\over 2}+1= \bbox[red, 2pt]{27\over 2}$$

解答:$$\textbf{(一) }  P(X\ge 0.5) = \int_{0.5}^1 f(x)\,dx =  \int_{0.5}^{1} 2(1 - x) \, dx = \left. \left[ 2x-x^2 \right] \right|_{0.5}^1 =\bbox[red, 2pt]{0.25} \\ \textbf{(二) }E[X]  = \int_{0}^{1} x \cdot 2(1 - x) \, dx =  \left. \left[ x^2-{2\over 3}x^3 \right] \right|_0^1 = \bbox[red, 2pt]{1\over 3} \\ \textbf{(三) }f(x\mid X\ge 0.5)= {f(x) \over P(X\ge 0.5)} ={2(1-x)\over 1/4}=8(1-x) \\\qquad \Rightarrow \bbox[red, 2pt]{f(x\mid X\ge 0/5) = \begin{cases} 8(1-x),& 0.5\le x\le 1\\ 0, & \mbox{otherwise.}\end{cases}}$$

解答:$$\textbf{(一) }   P(X=3) = \sum_{y=1}^{3} f_{XY}(3,y)  = f_{XY}(3,1) + f_{XY}(3,2) + f_{XY}(3,3)   \frac{1}{4} + 0 + \frac{1}{6}\\\qquad = \frac{3}{12} + \frac{2}{12} = \bbox[red, 2pt]{\frac{5}{12}} \\ \textbf{(二) } P(Y=1) = f_{XY}(1,1) + f_{XY}(2,1) + f_{XY}(3,1) = \frac{1}{4} + 0 + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \\ \qquad \Rightarrow  P(X=3 \mid Y=1) = \frac{f_{XY}(3,1)}{P(Y=1)} = \frac{1/4}{1/2} = \bbox[red, 2pt]{ \frac{1}{2}} \\\textbf{(三) } E[XY] = (1 \times 1)\left(\frac{1}{4}\right) + (3 \times 1)\left(\frac{1}{4}\right) + (2 \times 2)\left(\frac{1}{3}\right) + (3 \times 3)\left(\frac{1}{6}\right) = \bbox[red, 2pt]{23\over 6} \\ \textbf{(四) }\cases{P(X=1)=1/4\\ P(X=2)=1/3\\ P(X=3)=5/12} \Rightarrow E[X]= 1\cdot {1\over 4}+2\cdot {1\over 3}+3\cdot {5\over 12}={13\over 6} \\ \qquad \cases{P(Y=1)=1/2\\ P(Y=2)=1/3\\ P(Y=3)=1/6} \Rightarrow E[Y]= 1\cdot {1\over 2}+2\cdot {1\over 3}+3\cdot {1\over 6}={5\over 3} \\ \qquad E[X]\cdot E[Y] ={13\over 6}\cdot {5\over 3}={65\over 18} \ne E[XY] \Rightarrow Cov(X,Y) =E[XY]-E[X]E[Y]\ne 0 \\ \qquad \Rightarrow \bbox[red, 2pt]{否,兩者相關,因為Cov(X,Y) \ne 0}$$

解答:$$\textbf{(一) }第一球是紅球，剩下1紅、3白及4藍球\Rightarrow 抽中白球的機率={3\over 1+3+4}= \bbox[red, 2pt]{3\over 8} \\ \textbf{(二) }\cases{第一球是紅球且第二球是白球的機率= {2\over 9}\times {3\over 8}={6\over 72}\\ 第一球是白球且第二球是白球的機率= {3\over 9}\times {2\over 8}={6\over 72}\\ 第一球是藍球且第二球是白球的機率= {4\over 9}\times {3\over 8}={12\over 72}} \Rightarrow 合計為{24\over 72} = \bbox[red, 2pt]{1\over 3} \\ \textbf{(三) } {第一球是紅球且第二球是白球的機率\over 第二球是白球的機率} ={6/72 \over 1/3} =\bbox[red, 2pt]{1\over 4}$$

解答:$$\cases{P(X=1) =0.5\\ P(X=0)=0.5} \Rightarrow E[X] =1\times 0.5+0\times 0.5=0.5 \Rightarrow E[X^2]=1^2\times 0.5=0.5 \\ \Rightarrow Var(X)=E[X^2]-(E[X])^2 =0.5-0.25=0.25\Rightarrow \sigma _X =0.5 \\\textbf{(一) }Y=\sum_{i=1}^{10000} X_i \Rightarrow E[Y]=E \left[ \sum_{i=1}^{10000} X_i \right] =10000 \times 0.5=5000 \\\qquad \Rightarrow Var(Y)=10000\times Var(X)=2500 \Rightarrow \sigma_Y = \sqrt{2500}=50 \Rightarrow \bbox[red, 2pt]{Y \sim N(5000,2500)} \\ \textbf{(二) }P(Y\ge 5100) =P(Z\ge {5100-5000\over 50}) =P(Z\ge 2)=1-0.9772(查試題附表) \\\qquad = \bbox[red, 2pt]{0.0228}$$

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解題僅供參考，碩士班歷年試題及詳解