概率論考試題及答案大全

一、單項選擇題(每題2分,共10題)1.設(shè)隨機變量\(X\)服從參數(shù)為\(\lambda\)的泊松分布,且\(P(X=1)=P(X=2)\),則\(\lambda=\)（）A.0B.1C.2D.3答案：C解析:由泊松分布概率公式\(P(X=k)=\frac{\lambda^{k}e^{-\lambda}}{k!}\),已知\(P(X=1)=P(X=2)\),即\(\frac{\lambda^{1}e^{-\lambda}}{1!}=\frac{\lambda^{2}e^{-\lambda}}{2!}\),解得\(\lambda=2\)。2.若隨機變量\(X\)服從正態(tài)分布\(N(\mu,\sigma^{2})\),則隨著\(\sigma\)的增大,概率\(P(\vertX-\mu\vert\lt\sigma)\)（）A.單調(diào)增大B.單調(diào)減小C.保持不變D.增減不定答案：C解析:\(P(\vertX-\mu\vert\lt\sigma)=P(\mu-\sigma\ltX\lt\mu+\sigma)\),對于正態(tài)分布\(N(\mu,\sigma^{2})\),\(P(\mu-\sigma\ltX\lt\mu+\sigma)\)的值是固定的,約為\(0.6826\)。3.設(shè)\(X\)與\(Y\)相互獨立,且\(X\simN(0,1)\),\(Y\simN(1,1)\),則\(X+Y\)服從（）A.\(N(0,1)\)B.\(N(1,2)\)C.\(N(1,1)\)D.\(N(0,2)\)答案：B解析:若\(X\simN(\mu_{1},\sigma_{1}^{2})\),\(Y\simN(\mu_{2},\sigma_{2}^{2})\),且\(X\)與\(Y\)相互獨立,則\(X+Y\simN(\mu_{1}+\mu_{2},\sigma_{1}^{2}+\sigma_{2}^{2})\),這里\(\mu_{1}=0\),\(\mu_{2}=1\),\(\sigma_{1}^{2}=1\),\(\sigma_{2}^{2}=1\),所以\(X+Y\simN(1,2)\)。4.設(shè)\(X_{1},X_{2},\cdots,X_{n}\)是來自總體\(X\)的樣本,\(\overline{X}\)是樣本均值,\(S^{2}\)是樣本方差,則\(E(S^{2})\)等于（）A.\(E(X)\)B.\(D(X)\)C.\(E(X^{2})\)D.\(D(X^{2})\)答案：B解析:樣本方差\(S^{2}\)的數(shù)學(xué)期望\(E(S^{2})\)等于總體方差\(D(X)\)。5.設(shè)隨機變量\(X\)的分布函數(shù)為\(F(x)\),則\(P(a\ltX\leqb)=\)（）A.\(F(b)-F(a)\)B.\(F(b)-F(a-0)\)C.\(F(b+0)-F(a)\)D.\(F(b+0)-F(a-0)\)答案：B解析:根據(jù)分布函數(shù)的性質(zhì),\(P(a\ltX\leqb)=F(b)-F(a-0)\)。6.若事件\(A\)與\(B\)互斥,且\(P(A)\gt0\),\(P(B)\gt0\),則下列結(jié)論正確的是（）A.\(P(AB)=P(A)P(B)\)B.\(P(A\cupB)=P(A)+P(B)\)C.\(P(A\vertB)=P(A)\)D.\(P(B\vertA)\gt0\)答案：B解析:互斥事件\(A\)與\(B\)滿足\(P(A\cupB)=P(A)+P(B)\)。7.設(shè)隨機變量\(X\)的概率密度為\(f(x)\),且\(f(-x)=f(x)\),\(F(x)\)是\(X\)的分布函數(shù),則對于任意實數(shù)\(a\),有（）A.\(F(-a)=1-\int_{0}^{a}f(x)dx\)B.\(F(-a)=\frac{1}{2}-\int_{0}^{a}f(x)dx\)C.\(F(-a)=F(a)\)D.\(F(-a)=2F(a)-1\)答案：B解析:由\(f(-x)=f(x)\)可知\(X\)關(guān)于\(y\)軸對稱,\(F(-a)=\int_{-\infty}^{-a}f(x)dx\),\(\int_{-\infty}^{+\infty}f(x)dx=1\),\(\int_{-\infty}^{0}f(x)dx=\frac{1}{2}\),所以\(F(-a)=\frac{1}{2}-\int_{0}^{a}f(x)dx\)。8.設(shè)\(X\)服從參數(shù)為\(n,p\)的二項分布,且\(E(X)=1.6\),\(D(X)=1.28\),則\(n\),\(p\)的值分別為()A.\(n=8\),\(p=0.2\)B.\(n=4\),\(p=0.4\)C.\(n=5\),\(p=0.32\)D.\(n=6\),\(p=0.27\)答案：A解析:二項分布\(E(X)=np\),\(D(X)=np(1-p)\),由\(np=1.6\),\(np(1-p)=1.28\),解得\(n=8\),\(p=0.2\)。9.設(shè)\(X\),\(Y\)為隨機變量,\(E(X)=E(Y)=1\),\(D(X)=4\),\(D(Y)=9\),\(\rho_{XY}=-\frac{1}{6}\),則\(E(X-2Y+1)\),\(D(X-2Y+1)\)分別為()A.\(-1\),\(28\)B.\(-1\),\(21\)C.\(0\),\(28\)D.\(0\),\(21\)答案：A解析:\(E(X-2Y+1)=E(X)-2E(Y)+1=1-2\times1+1=-1\);\(D(X-2Y+1)=D(X)+4D(Y)-4\rho_{XY}\sqrt{D(X)D(Y)}=4+4\times9-4\times(-\frac{1}{6})\times\sqrt{4\times9}=28\)。10.設(shè)\(X_{1},X_{2},\cdots,X_{n}\)是來自總體\(N(\mu,\sigma^{2})\)的樣本,\(\overline{X}\),\(S^{2}\)分別為樣本均值和樣本方差,則（）A.\(\overline{X}\simN(\mu,\frac{\sigma^{2}}{n})\)B.\(\frac{(n-1)S^{2}}{\sigma^{2}}\sim\chi^{2}(n-1)\)C.\(\frac{\overline{X}-\mu}{S}\sqrt{n}\simt(n)\)D.\(\frac{\overline{X}-\mu}{\sigma}\simN(0,1)\)答案：B解析:根據(jù)抽樣分布的性質(zhì),\(\frac{(n-1)S^{2}}{\sigma^{2}}\sim\chi^{2}(n-1)\)。二、多項選擇題(每題2分,共10題)1.下列函數(shù)中,可以作為隨機變量分布函數(shù)的是（）A.\(F(x)=\left\{\begin{array}{ll}0,&x\lt0\\\frac{1}{2},&0\leqx\lt1\\1,&x\geq1\end{array}\right.\)B.\(F(x)=\left\{\begin{array}{ll}0,&x\lt0\\\sinx,&0\leqx\lt\frac{\pi}{2}\\1,&x\geq\frac{\pi}{2}\end{array}\right.\)C.\(F(x)=\left\{\begin{array}{ll}0,&x\lt0\\\frac{x}{1+x},&x\geq0\end{array}\right.\)D.\(F(x)=\left\{\begin{array}{ll}0,&x\lt0\\1-e^{-x},&x\geq0\end{array}\right.\)答案:CD解析:分布函數(shù)需滿足右連續(xù)性等性質(zhì),C和D選項滿足分布函數(shù)的性質(zhì)。2.若隨機變量\(X\)與\(Y\)相互獨立,則（）A.\(E(XY)=E(X)E(Y)\)B.\(D(X+Y)=D(X)+D(Y)\)C.\(P(XY)=P(X)P(Y)\)D.\(Cov(X,Y)=0\)答案:ABD解析:相互獨立的隨機變量\(X\)與\(Y\)滿足\(E(XY)=E(X)E(Y)\),\(D(X+Y)=D(X)+D(Y)\),\(Cov(X,Y)=E(XY)-E(X)E(Y)=0\)。3.設(shè)\(X\simN(\mu,\sigma^{2})\),則（）A.\(P(X\gt\mu)=0.5\)B.\(P(X\lt\mu-k\sigma)=P(X\gt\mu+k\sigma)\)C.\(P(\vertX-\mu\vert\ltk\sigma)=2\varPhi(k)-1\)D.\(P(X\lt\mu+k\sigma)=1-\varPhi(k)\)答案:ABC解析:正態(tài)分布\(N(\mu,\sigma^{2})\)關(guān)于\(x=\mu\)對稱,所以\(P(X\gt\mu)=0.5\),\(P(X\lt\mu-k\sigma)=P(X\gt\mu+k\sigma)\),\(P(\vertX-\mu\vert\ltk\sigma)=2\varPhi(k)-1\)。4.設(shè)\(X_{1},X_{2},\cdots,X_{n}\)是來自總體\(X\)的樣本,\(\overline{X}\)為樣本均值,\(S^{2}\)為樣本方差,則有（）A.\(\sum_{i=1}^{n}(X_{i}-\overline{X})^{2}\sim\chi^{2}(n-1)\)B.\(\frac{\overline{X}-\mu}{S/\sqrt{n}}\simt(n-1)\)C.\(\frac{(n-1)S^{2}}{\sigma^{2}}\sim\chi^{2}(n)\)D.\(\overline{X}\simN(\mu,\frac{\sigma^{2}}{n})\)答案:BD解析:\(\frac{\overline{X}-\mu}{S/\sqrt{n}}\simt(n-1)\),\(\overline{X}\simN(\mu,\frac{\sigma^{2}}{n})\)。5.設(shè)\(A\),\(B\)為隨機事件,則下列等式成立的有（)A.\(P(A\cupB)=P(A)+P(B)\)(\(A\)與\(B\)互斥)B.\(P(AB)=P(A)P(B)\)(\(A\)與\(B\)獨立)C.\(P(A-B)=P(A)-P(AB)\)D.\(P(A\cupB)=P(A)+P(B)-P(AB)\)答案:ABCD解析:根據(jù)概率的基本公式和性質(zhì),這些等式都是正確的。6.設(shè)隨機變量\(X\)的概率分布為\(P(X=k)=\frac{C}{k!}\),\(k=0,1,2,\cdots\),則（）A.\(C=e\)B.\(P(X\geq1)=1-e^{-1}\)C.\(E(X)=1\)D.\(D(X)=1\)答案:ABCD解析:由概率分布的性質(zhì)\(\sum_{k=0}^{\infty}P(X=k)=1\)可求出\(C=e\),進而可求出其他期望和方差等。7.設(shè)\(X\simU(0,2)\),則（）A.\(E(X)=1\)B.\(D(X)=\frac{1}{3}\)C.\(P(X\geq1)=\frac{1}{2}\)D.\(F(x)=\left\{\begin{array}{ll}0,&x\lt0\\\frac{x}{2},&0\leqx\lt2\\1,&x\geq2\end{array}\right.\)答案:ABCD解析:均勻分布\(U(a,b)\)的期望、方差、概率等可根據(jù)公式計算。8.下列說法正確的是()A.若\(X\)服從指數(shù)分布,則\(E(X)=\frac{1}{\lambda}\),\(D(X)=\frac{1}{\lambda^{2}}\)B.若\(X\sim\chi^{2}(n)\),則\(E(X)=n\),\(D(X)=2n\)C.若\(X\simt(n)\),則\(E(X)=0\),\(D(X)=\frac{n}{n-2}(n\gt2)\)D.若\(X\simF(m,n)\),則\(E(X)=\frac{n}{n-2}(n\gt2)\)答案:ABC解析:根據(jù)常見分布的期望和方差公式判斷。9.設(shè)\(X\),\(Y\)是隨機變量,且\(E(XY)\)存在,則（)A.\(E(XY)=E(X)E(Y)\)(\(X\)與\(Y\)獨立)B.\(Cov(X,Y)=E(XY)-E(X)E(Y)\)C.\(D(X+Y)=D(X)+D(Y)+2Cov(X,Y)\)D.\(D(X-Y)=D(X)+D(Y)-2Cov(X,Y)\)答案:BCD解析:獨立時\(E(XY)=E(X)E(Y)\),協(xié)方差和方差的運算公式滿足B.C.D選項。