2025年考研數(shù)學《概率論》沖刺押題試卷考試時間：______分鐘總分：______分姓名：______一、選擇題：本大題共5小題，每小題4分，共20分。在每小題給出的四個選項中，只有一項是符合題目要求的。1.設(shè)事件A與B互斥，且P(A)>0，P(B)>0，則下列結(jié)論中一定正確的是（）（A）A與B獨立（B）A與B不獨立（C）A與B互為對立事件（D）A與B可能相互獨立2.設(shè)隨機變量X的分布函數(shù)為F(x)，則P(X≤0)等于（）（A）0（B）F(0)-（C）F(0)（D）1-F(0)3.設(shè)隨機變量X與Y獨立同分布，且均服從參數(shù)為λ的泊松分布，則E[X(Y-X)]等于（）（A）λ（B）λ^2（C）λ^3（D）04.設(shè)二維隨機變量(X,Y)的聯(lián)合概率密度函數(shù)為f(x,y)=\begin{cases}cxy,&0≤x≤1,0≤y≤x\\0,&其他\end{cases}，則常數(shù)c等于（）（A）1（B）2（C）-1（D）-25.設(shè)隨機變量X和Y的期望E(X)=1，E(Y)=2，方差Var(X)=1，Var(Y)=4，且Cov(X,Y)=1，則X+2Y的方差Var(X+2Y)等于（）（A）9（B）11（C）13（D）15二、填空題：本大題共4小題，每小題4分，共16分。6.設(shè)A，B，C為三個事件，且P(A)=P(B)=P(C)=1/4，P(AB)=P(AC)=P(BC)=1/8，P(ABC)=1/16，則P(A∪B∪C)等于________。7.設(shè)隨機變量X服從參數(shù)為n=10，p=0.2的二項分布，則P(X=3)等于________。8.設(shè)隨機變量X的密度函數(shù)為f(x)=\begin{cases}2x,&0≤x≤1\\0,&其他\end{cases}，則P(0.5<X<1)等于________。9.設(shè)隨機變量X和Y的協(xié)方差Cov(X,Y)=3，X的方差Var(X)=4，Y的方差Var(Y)=9，則X和Y的相關(guān)系數(shù)ρ_{XY}等于________。三、解答題：本大題共6小題，共74分。解答應寫出文字說明、證明過程或演算步驟。10.（本題滿分12分）設(shè)隨機變量X和Y相互獨立，X的密度函數(shù)為f_X(x)=\begin{cases}1,&0≤x≤1\\0,&其他\end{cases}，Y的密度函數(shù)為f_Y(y)=\begin{cases}e^{-y},&y>0\\0,&y≤0\end{cases}。求隨機變量Z=2X+Y的密度函數(shù)。11.（本題滿分12分）設(shè)隨機變量X和Y相互獨立，且都服從標準正態(tài)分布N(0,1)。求隨機變量U=X^2+Y^2和V=X^2-Y^2的協(xié)方差Cov(U,V)。12.（本題滿分12分）設(shè)隨機變量X的分布函數(shù)為F(x)=\begin{cases}0,&x<0\\x^2,&0≤x≤1\\1,&x>1\end{cases}。求隨機變量X的期望E(X)和方差Var(X)。13.（本題滿分13分）設(shè)二維隨機變量(X,Y)的聯(lián)合分布律如下表所示（其中a為常數(shù)）：Y\X|0|1|----|------|------|0|0.1|a|1|0.3|0.2|（1）求常數(shù)a的值；（2）求隨機變量X的邊緣分布律；（3）判斷X和Y是否相互獨立。14.（本題滿分13分）從一批產(chǎn)品中重復抽取產(chǎn)品進行檢驗，每次抽取一件，假設(shè)抽到次品的概率為p（0<p<1），連續(xù)抽到第2件次品之前，最多只抽到3件正品。求p的矩估計量。15.（本題滿分10分）設(shè)總體X服從正態(tài)分布N(μ,σ^2)，其中μ未知，σ^2已知。設(shè)X_1，X_2，...，X_n是從該總體中抽取的一個樣本。證明：統(tǒng)計量\bar{X}=\frac{1}{n}\sum_{i=1}^nX_i是總體均值μ的無偏估計量。試卷答案一、選擇題1.B2.C3.D4.B5.C二、填空題6.3/47.120/10248.3/49.1/3三、解答題10.解：因為X和Y相互獨立，所以Z=2X+Y的密度函數(shù)f_Z(z)可以通過卷積公式計算：f_Z(z)=\int_{-\infty}^{+\infty}f_X(x)f_Y(z-2x)dx由于f_X(x)在[0,1]上為1，其他地方為0，f_Y(y)在y>0時為e^{-y}，其他地方為0，所以需要考慮z-2x的取值范圍。當z<0時，f_Y(z-2x)永遠不為正，所以f_Z(z)=0。當0≤z≤2時，需要0≤x≤1且z-2x>0，即0≤x≤z/2，所以：f_Z(z)=\int_{0}^{z/2}1*e^{-(z-2x)}dx=e^{-z}\int_{0}^{z/2}e^{2x}dx=e^{-z}[e^{2x}/2]_{0}^{z/2}=e^{-z}[e^z/2-1/2]=(1/2)-(e^{-z}/2)當z>2時，需要0≤x≤1且z-2x>0，即(z/2)≤x≤1，所以：f_Z(z)=\int_{z/2}^{1}1*e^{-(z-2x)}dx=e^{-z}\int_{z/2}^{1}e^{2x}dx=e^{-z}[e^{2x}/2]_{z/2}^{1}=e^{-z}[e^2/2-e^z/(2e^z)]=e^{-z}[e^2/2-1/(2e^2)]=(e^2-1)/(2e^2)*e^{-z}綜上，Z的密度函數(shù)為：f_Z(z)=\begin{cases}0,&z<0\\(1/2)-(e^{-z}/2),&0≤z≤2\\(e^2-1)/(2e^2)*e^{-z},&z>2\end{cases}11.解：因為X和Y相互獨立且都服從N(0,1)，所以X^2和Y^2都服從χ^2(1)分布，且X^2和Y^2相互獨立。Cov(U,V)=Cov(X^2+Y^2,X^2-Y^2)=Cov(X^2,X^2)-Cov(X^2,Y^2)由于X^2和Y^2獨立，Cov(X^2,Y^2)=0。Cov(X^2,X^2)=Var(X^2)E(X^2)=1(因為X~N(0,1))E(X^4)=3(因為X~N(0,1))Var(X^2)=E(X^4)-[E(X^2)]^2=3-1^2=2所以Cov(U,V)=2-0=212.解：E(X)=\int_{-\infty}^{+\infty}x*f(x)dx=\int_{0}^{1}x*2xdx=2\int_{0}^{1}x^2dx=2*[x^3/3]_{0}^{1}=2/3E(X^2)=\int_{-\infty}^{+\infty}x^2*f(x)dx=\int_{0}^{1}x^2*2xdx=2\int_{0}^{1}x^3dx=2*[x^4/4]_{0}^{1}=1/2Var(X)=E(X^2)-[E(X)]^2=1/2-(2/3)^2=1/2-4/9=1/1813.解：（1）由邊緣分布律性質(zhì)，P(X=0)=P(X=0,Y=0)+P(X=0,Y=1)=0.1+0.3=0.4P(X=1)=P(X=1,Y=0)+P(X=1,Y=1)=a+0.2=1-P(X=0)=1-0.4=0.6所以a+0.2=0.6，得a=0.4。（2）X的邊緣分布律為：X|0|1|P|0.4|0.6|（3）判斷獨立性，需要看P(X=1,Y=1)是否等于P(X=1)P(Y=1)。P(X=1,Y=1)=0.2P(X=1)=0.6，P(Y=1)=P(Y=1,X=0)+P(Y=1,X=1)=0.3+0.2=0.5P(X=1)P(Y=1)=0.6*0.5=0.3因為P(X=1,Y=1)=0.2≠0.3=P(X=1)P(Y=1)，所以X和Y不相互獨立。14.解：設(shè)X為抽到第2件次品之前抽到的正品件數(shù)，則X的可能取值為0,1,2,3。X=0表示第一次抽到次品，P(X=0)=p。X=1表示第一次抽到正品，第二次抽到次品，P(X=1)=(1-p)p。X=2表示前兩次抽到正品，第三次抽到次品，P(X=2)=(1-p)^2*p。X=3表示前三次抽到正品，第四次抽到次品，P(X=3)=(1-p)^3*p。X≥4表示前四件都抽到正品，第五件抽到次品，P(X≥4)=(1-p)^4。題目條件“連續(xù)抽到第2件次品之前，最多只抽到3件正品”意味著X的可能取值為0,1,2,3。因此，樣本空間為S={0,1,2,3}，對應的概率函數(shù)為：P(X=k)=\begin{cases}p,&k=0\\(1-p)p,&k=1\\(1-p)^2p,&k=2\\(1-p)^3p,&k=3\end{cases}總概率為1，驗證：(1-p)^4=1-p-(1-p)p-(1-p)^2p-(1-p)^3p=p^3，成立?？傮w中次品的概率為p，抽到k件正品后再抽到次品的概率為p，所以P(X=k)是k次正品后抽到第1個次品的概率。E(X)=\sum_{k=0}^3k*P(X=k)=0*p+1*(1-p)p+2*(1-p)^2p+3*(1-p)^3p=p(1-p+2(1-p)^2+3(1-p)^3)E(X)=p(1-p+2(1-2p+p^2)+3(1-3p+3p^2-p^3))E(X)=p(1-p+2-4p+2p^2+3-9p+9p^2-3p^3)E(X)=p(6-14p+11p^2-3p^3)E(X)=6p-14p^2+11p^3-3p^4令g(p)=6p-14p^2+11p^3-3p^4，求g'(p)：g'(p)=6-28p+33p^2-12p^3=3p^2(11-12p)+2(3-14p)=3p^2(11-12p)+2(3-14p)E(X)=\sum_{k=0}^3k*P(X=k)=\sum_{k=1}^3k*\binom{k}{k}*p^k*(1-p)^{k-1}=p\sum_{k=1}^3k*\binom{k-1}{1}*p^{k-1}*(1-p)^{k-2}=p\sum_{j=1}^2(j+1)*\binom{j}{1}*p^j*(1-p)^{j-1}=p\sum_{j=1}^2(j+1)*j*p^j*(1-p)^{j-1}=p\sum_{j=1}^2(j+1)*j*[p(1-p)^{j-1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]=p\sum_{j=1}^2(j+1)*j*[(1-p)^j-(1-p)^{j+1}]