的條件概率．P(B|A)讀作A發(fā)生的條件下B發(fā)生的概率．公式計算P(B|A)就沒有意義了，所以條件概率計算必須在P(A)(3)如果B和C是兩個互斥事件，則P(B∪C|A)=P(BB包含的基本事件數(shù)n(AB)，則P(B|A．第二次拋出的點數(shù)是奇數(shù)的概率為(A.B.C.D.則P(B|AX0<X<11≤X<22≤X<3X≥3Y134A.B.C.D.【詳解】由概率的加法公式可得P(X≥1(=1-P(X<1)=0.6,P(1≤X<3)=0.9-0.4=0.5，故P(Y≤4|X≥1(=P(X<3|X≥1)===．3.(24-25高二下·重慶·期末)記A為事件A的對立事件，且P(AP(B(=,則P(A+B(=()A.B.C.D.【詳解】由題意得P(A)=，P(AB)=P(A|B(P(B(=×=，因為P(B)=P(A∩B(+P(A—∩B(，得P(A∩B(=P(B)-P(A—∩B(=-=，則P(A+B(=P(A(+P(B(-P(AB(=+-=.,則P(B(等于()A.B.C.D.所以P(B(=1-P(B-(=1-=.A.B.C.D. 則P(AP(AB，則P(B|A(==×=，()A.B.C.D.則A=A1UA1A2，事件A1,A1A2互斥，所以P(A)=P(A1)+P(AA2)=P(A1)+P(A)P(A2|A)=+×=,A.事件A1，A2為對立事件B.PC.事件B，C為獨立事件D.P(C|AA1A正確. ,積事件BC為兩個紅色球，則P(BC,可知P(B(.P(C(=×=≠P(BC(，根據(jù)條件概率公式可知P(C|A2(===，所以D正確.A.PB.PC.PD.P【詳解】對于A，“P(A(=,P(B(=,:P(B-(=1-P(B(=1-=，:P(AUB-(=P(A(+P(B-(-P(AB-(=+-P(AB-(=，解得P(A，故A錯誤；對于B，“P(AB-(+P(AB(=P(A(，:+P(AB(=，解得P(AB(=，:PP(AB故B正確；C.根據(jù)分步計數(shù)原理有n(A(=AA=20，所以P(A故A正確； 對于B，n(AB(=A則P(AB故B錯誤；對于D，P(B|A故D正確.則P(A(=0.8×(1-0.9(+(1-0.8(×0.9=0.26，P(AB(=0.8×(1-0.9(=0.08，則P(B|A11.(24-25高二下·上海·期,則P(AB(=．所以P(AB(=P(B|A(P(A(=×=.比賽進行3局且甲班獲勝的概率為P(3局且甲勝)=2...=,(1)0.7192該部門招工的淘汰率為P(B-(=1-P(B(=1(1)(1)記事件A:第一次取出的是一次性手套，事件B:第二次取出的是一次則P(BP(AB，由條件概率公式可得P(A|B根據(jù)題意可得P(C (1)平均課外閱讀時間32小時/月；所以該校學(xué)生課外閱讀時間位于區(qū)間[30,60((單位：小時/月)停車時間/分鐘(0,15[甲a 38a 8乙b 3 6(2)X+Y=30有以下情況：所以P(X+Y=30(=++=,所以P(X=Y(1) 故P(B|A(1)求p；(1)p=則P(A|B故甲在4輪比賽之內(nèi)獲勝的概率P=P(C(+P(D(+P(E(=++=.(1)0.84;故P(B)=P(BC)+P(BD)=0.0576+0.0384=0.096；故比賽在第四局結(jié)束條件下甲最終獲勝的概率P； P1.概率的乘法公式：對于任意兩個事件A、B，若P(A)>0，則P(AB)=P(A)P(B|A(.A2n=Ω,且P(Ai)>0，i=1,2,3...,n，則對于任意的事件B?Ω,有PP(B|Ai(.A2n=Ω,且P(Ai)>0，i=1,2,3..率)核心作用體現(xiàn)先驗概率到后驗概率的更新則P(A)=0.25，P(B)=0.3，P(C)=0.45，P(D|A(=0.06，P(D|A(=0.06，P(D|B(=P(D|C(=0.08任取一個零件是次品的概率:P(D(=P(A(P(D|A(+P(B(P(D|B(+P(C(P(D|C(=0.25×0.06+ A.B.C.D.則P(B(=P(A0(P(B|A0(+P(A1(P(B|A1(+P(A2(P(B|A2(=×1+×+×=．A.0.4B.0.5則P(B|A23.(2025·江西·模擬預(yù)測)兒童牙齒是否健康與早晚是否都刷牙有關(guān).據(jù)調(diào)查，某幼兒園大約有 ()A.B.C.D.則P(A)=1-0.3=0.7,P(AB)=0.39，所以P(B∣A)===.A.PB.PC.PD.P(B|A-)=由概率的乘法公式可得P(AB(=P(A(P(B|A)=×=，故A錯誤.由全概率公式可得P(B(=P(AB(+P(A-B(=+=，故B錯誤.對于D，由B項分析，已得P(B故D正確.優(yōu)勢R，在B發(fā)生的條件下A的優(yōu)勢R則() 已知R由條件概率公式P(A|BP(A由列聯(lián)表可知P(AB(=,P(AB—(=,P(AB(=,P(AB(=，P(B|A(==B.他連續(xù)兩天都去A餐廳的概率為C.他連續(xù)兩天都不去A餐廳的概率為設(shè)事件D={第二天到餐廳A用餐}，則其對立事件D—={對于A，P(D(=P(D|C(P(C(+P(D|C-(P(C對于B，P(CD(=P(D|C(P(C(=，故B錯誤；-P(DC-(=P(C-(-P(D|C-(P故C錯誤；對于D，P(C|D故D正確.A.A1,A2是對立事件B.P(B2∣AC.P(BD.P(A1∣B1(=所以A1與A2是對立事件，故A正確；根據(jù)條件概率的含義得P(B2|A2)==，P(B1|A2)==，故B正確；對于C：由題得P(A1)=P(A2)==，計算得P(B1|A1P(B1)=P(A1)P(B1|A1)+P(A2)P(B1|A2)，即P(B1)=×+×=≠,故C錯誤； ∴P(B|A(=，P(B|A(=，P(A(=，∴P(B|A(=1-P(B|A(=1-=，P(A(=，對于A選項，∵P(B|AP(AB(=P(B|A(P(A，∴P(A(P(B(=×=≠P(AB(，∵P(B|A(=，∴P(B|A(=1-P(B|A(=，∴P(AB(=P(A(P(B|A(=×=，根據(jù)條件概率公式可得，P(AB(=P(A(P(B|A(=×=，再根據(jù)貝葉斯公式可得，P(A故D錯誤.所以P(A1(=P，則P(B-|Ai，所以第2次投籃人是甲的概率為P=P+P由題知P(B2|AP(B2|A2(=0，P(B2|AP(B2|A，又P(A1(=P(A2(=P(A3(=P(A所以P(BP(Ai(P(B2|Ai又P(A4|B 32.(24-25高二下·貴州貴陽·階段練習(xí))把若干個紅球和白球(除顏色外沒有其他差異)放進甲、(1)則P(D(=P(A(.P(D|A(+P(B(.P(D|B(+P(C(.P(D|C(分別記A|D、B|D、C|D為H,I,J，則P(E(=P(H(.P(E|H(+P(I(.P(E|I(+P(J(.P(E|J( 且P(A(=P(A-(=，所以P(X=0(=P(A(.P(X=0|A(+P(A-(.P(X=0|A-(P(X=1(=P(A(.P(X=1|A(+P(A-(.P(X=1|A-(P(X=2(=P(A(.P(X=2|A(+P(A-(.P(X=2|A-(所以X的分布列為X012P 6 2 3故X的均值為E(X(=0×+1×+2×=．(2)設(shè)甲獲得比賽勝利為事件B，則P(B|X=0(=0,P(B|X=1(=P(B(,P(B|X=2(=1．由(1)知，P(XP(XP(X得P(B(=P(X=0(P(B|X=0(+P(X=1(P(B|X=1(+P(X=2(P(B|X=2(=×0+P(B(+ 解得P(B(=． P(MN(==,P(M(==,P(N(=×+×==,所以P(N|MP(M|N由題意可得P(A)=0.3,P(B)=0.4,P(C)=0.3,P(E|A)=0.05,P(E|B)=0.04，P(E|C)=0.03，由全概率公式可得P(E)=P(A).P(E|A)+P(B).P(E|B)+P(C).P(E|C)=0.3×0.05+0.4×所以P(C|E概率.(1)所以P(A從甲機器生產(chǎn)的產(chǎn)品中各任取1件產(chǎn)品為合格產(chǎn)品為從乙機器生產(chǎn)的產(chǎn)品中各任取1件產(chǎn)品為合格產(chǎn)品為事件B2，—B—所以P(B(=P(BB2B3(+P(B1BB3(+P(B1B2B(根據(jù)已知條件有：P(C1(=P(C2(=P(CP(D(=P(D|C1(P(C1(+P(D|C2(P(C2(+P(D|C3(P(C3(甲乙丙(1)0.014B2P(A(=P(B1(P(A|B1(+P(B2(P(A|B2(+P(B3(P(A|B3(P(P(A∣B1(P(B1(=P(A( X0123P(1)求m與p的比值；2>0)；Y12345Pp1p2(1)隨機變量X表示一個家庭安裝智能燈泡的數(shù)量，其分布列概率和等于1，m-p=0,m=p；=P(Y=1(=P=P-P解得p計算P(S≥4|X=k)：，故P(S≥4|X=1)=P=P+PY2Y2≥4).計算P(S<4)=P(S≤3(：,Y2Y2+Y3，需P(S≥4).計算P(S<4)=P(S≤3)：故P則：P(A(=P(SP(S≥4|X=k(P(X=k(設(shè)函數(shù)f(p)=P(A)，p∈(0,1).求導(dǎo)得fI(p單調(diào)遞增，(q,1(內(nèi)gI(p(<0，g(f(p(單調(diào)遞增，(r,1(內(nèi)g(p(<0，f(p(單調(diào)減，+- P(AB(=P(A(+P(B(-P(B∪A(=P(A(P(B(，所以A,B獨立，所以P(A|B(=P(A(=,P(A-|B(=，(2)如果主持人不知道內(nèi)情，P(A(=P(B于是，P(A|B(===P(B|A(=,P(A-|B(=，說明換箱子與不換箱子中獎概率都是.(3)如果主持人知道內(nèi)情的概率為p，事件E表示主==P(A|BE(P(E|B(+P(A|BE-(P(E-|B(，又P(B|E(=1,P(B|E-(=,P(A|BE(=,P(A|BE-(=，設(shè)q=1-p， P(E(P(B|E(p3pP(E|B(=P(E(P(B|E(+P(E-( P(E(P(B|E(p3p說明不換箱子不中獎的概率.P(A-|B(=1-P(A|B(=.于前一步或前幾步的狀態(tài).這時可以通過建立遞推關(guān)系式來描述概率.通過全概率分解得到形如Pn=a.Pn-1+b(1-Pn-1(的遞推式.照“沒有連續(xù)相同的選項”猜答案．設(shè)其答對第n題的概率是Pn．則下列說法正確的是 () A.P(猜對第n+1題|猜對第n題B.P(猜對第n+1題|猜錯第n題)=設(shè)每道題的選項為A,B,C，不妨設(shè)第n題的正確答案為A，則P(An+1|An(=，故A選項正確；則在該同學(xué)第n題答錯的條件下，該同學(xué)這兩道題的選項可能為(B,A(,(B,C((C,A(,(C,B(，其中第n+1題答對的是(C,B(，故P(An+1|A(=，故B選項錯誤；容易得出Pn+1=Pn+(1-Pn(=Pn+，則Pn+1-=Pn-，全部猜對的概率P(A1(P(A2|A1(P(A3|A2(…P(A20|A19(=×19，故D錯誤.D.球第n(n∈N*(次傳出后恰好傳給丙的概率為+-n-1則n-1設(shè)pn+1+tnpn所以pnn-1，所以pnn-1.所以pin.n.nn設(shè)P(An(=pn，依題可知P(Bn(=1-pn.則P(An+1(=P(AnAn+1(+P(BnAn+1(=P(An(P(An+1|An(+P(Bn(P(An+1|Bn(. 即pn=×n-1+.所以第n次聽寫的人是甲的概率為pn=×n-1+.套餐的概率為Pn(n∈N*(，則Pn=【詳解】設(shè)第n天選擇米飯?zhí)撞蜑槭录嗀n，則第n天選擇面食套餐為Bn，則P(An(=Pn,P(Bn(=1-Pn，由題意可知P(An+1|An(=,P(An+1|Bn(=，由全概率公式得Pn+1=P(An+1(=P(An(P(An+1|An(+P(Bn)P(An+1|Bn(=Pn+(1-Pn(，所以Pn-=×(-n-1，則Pn=×(-n-1+.的概率為Pn.-2滿足的關(guān)系式；(1)P1=(2)Pn=Pn-1+Pn-2*)設(shè)Pn+λ=-(Pn-1+λ(，解得λ=-，P1-=-≠0.所以Pn-=(--n-1，所以Pn=(--n-1+，:最小值為P1=；(i)證明：{P(n+1(-P(n({為等比數(shù)列．( (0),P/(1),P/(2)， 于是P(n+2)=P(An)P(An+2|An)+P(An+1)P(An+2|An+1)=P(n)+P(n+1)，則P(n+2)-P(n+1)=-[P(n+1)-P(n)]，又P(1)=,P(2)=+×=,P(2)-P(1)=所以{P(n+1(-P(n({是首項為,公比為-的等比數(shù)列．(ii)由(i)得，當(dāng)n≥2時，P(n)-P(n-1)=×(-n-2=(-n，累加得P(n)-P(1)=(-2+(-3+…+(-n=-，因此P(n)=|1-(-n+1|，當(dāng)n≥5,n為奇數(shù)時，P(n)=n+1單調(diào)遞增，且P(n)<，當(dāng)n≥5且n為偶數(shù)時，P(n(=|1+n+1|單調(diào)遞減，且P(n(>，(1)根據(jù)題意得，P(A1(=,P(A(=,P(A2|A1(=,P(A2|A(=1-=，所以P(A2(=P(A1(P(A2|A1(+P(A(P(A2|A(=×+×=．(2)設(shè)An表示第n天選擇綠豆湯，則Pn=P(An(,P(A(=1-Pn，根據(jù)題意得，P(An+1|An(=,P(An+1|A(=1-=，由全概率公式得，P(An+1(=P(An(P(An+1|An(+P(A(P(An+1|A(=Pn+(1-Pn(=-Pn+，+1-=-Pn-，又P1-=≠0，則+×(-n-1>，即(-n-1>(n=1,2,…,10)，當(dāng)n=1,3,5,7,9時，考慮(-n-1=n-1>的解，(2)記該同學(xué)第n天選擇米飯?zhí)撞偷母怕蕿镻n；(1)；— 根據(jù)題意P(AP(A2|AP(A由全概率公式得：P(A2(=P(A1(P(A2|A1(+P(A(P(A2|A(=×+×=.n根據(jù)題意，P(An+1|AnP(An+1=P(An+1(=P(An(P(An+1|An(+P(A(P(An+1|A(=Pn+(1-Pn(=-Pn+因此Pn+1-=-Pn-，∵P1-=≠0.②：根據(jù)①可得Pn-=-n-1(n≥2(，此時{Pn{是單調(diào)遞減數(shù)列，所以Pn的最大值為P3=+因此m≥,則m的取值范圍是m∈,+∞(.(1)0.5;(2)證明見解析，pn-=(--n-1，qn+pn-=-n.設(shè)ti,j為第1則P(t0,0(=0.4×0.5=0.2,P(t0,1(=0.4×0設(shè)第1分鐘時，貓和老鼠所在房間號之和為X，則P(X=1(=P(t0,1(+P(t1,0(=0.5，上一分鐘在1號房間，轉(zhuǎn)移到0號房間的概率為(1-pn-1(，所以pn=pn-1+(1-pn-1(=-pn-1，則pn-=-pn-1-，上一分鐘貓和老鼠都在1號房間，老鼠轉(zhuǎn)移到0號房間的概率為(1-pn-1((1-qn-1(；n-1(1-pn-1(..所以qn=(1-pn-1((1-qn-1(+pn-1(1-qn-1(.+qn-1(1-pn-1(.，因為pn=-n+，所以qn=1--n-1+-=--n-1-，n---n-1=-qn-1---n-2，又因為q1--=-，所以數(shù)列{qn---n-1(L{是首項為-，公比為-的等比數(shù)列，n---n-1=--n-1，q0=0滿足上式也滿足題意，則qn=+-n-1+n+pn-=+-n-1+-n+-n+-=-n，(1)0.45 (3)Pk=-×(-k-1則P(A2(=P(A1(P(A2|A1(+P(B1(P(A2|B1(=0.5×0.3+0.5×(1-0.4(=0.45;(2)X可能的取值為0,1,2，大為P(Xn≥3，n≥an+1,an≥an-1≥,所以當(dāng)n=3,ann=4,an故P(X=1(的最大值為，此時n=3或4.則P(Ai(=1-Pi，則有P(Bi+1(=P(Bi(P(Bi+1|Bi(+P(Ai(P(Bi+1|Ai(=0.4Pi+(1-0.3((1-Pi(=0.7-0.3Pi，則Pi-=-×(-i-1，故Pk=-×(-k-1.50.(24-25高二下·山東泰安·期末)為備戰(zhàn)全國機器人大賽，某高校機器人甲隊和乙隊進行果互不影響.(1)；(2)P(B|An-1)=，P(B|An)=；(1)P(X≥1)=P(X=1)+P(X=3)=C2×+3=，則剩余2局的贏局?jǐn)?shù)Y～B(2,，總分滿足k+Y≥n所以An-1對應(yīng)(n-1)+Y≥n+1，即Y≥2，又Y≤2，故P(B|An-1)=×=，對于An對應(yīng)n+Y≥n+1，即Y≥1，又Y≤2，所以P(B|An)=1-P(Y=0)=1-(1-2=；(3)由全概率公式得Pn+1=C--11pn-1(1-p)np2+C-1pn(1-p)n-1[1-(1-p)2]+[Pn-C-1pn(1-p)n-1]=Pn+C--11pn-1(1-p)np2-C-1pn(1-p)n-1(1-p)2=Pn+C--11pn+1(1-p)n-C-1pn(1-p)n+1=Pn+C-1pn(1-p)n(2p-1)，∴Pn+1-Pn=C-1pn(1-p)n(2p-1)，+1-Pn>0，∵Pn+1-Pn>0，∴Pn+2-Pn+1<Pn+1-Pn， ()A.B.C.D.則P(D(=P(A(P(D|A(+P(B(P(D|B(+P(C(.P(D|C(P(CD(=P(C(P(D|C(=，則他騎共享單車的概率為P(C|DA.B.C.D. 2根據(jù)全概率公式：P(A)=P(B1)P(A|B1)+P(B2)P(A|B2)，54.(24-25高二下·安徽安慶·期末)已知事件A，B滿足P(AP(BP(A，則P(A-|B(=()A.B.C.D.P(A+B-(=P(A(+P(B-(-P(AB-(=，所以P(AB(=P(A(-P(AB-(=-=，故P(A-B(=P(B(-P(AB(=-=， 由條件概率公式得P.55.(24-25高二下·天津河西·期末)已知隨機事件A，B發(fā)生的概率分別為P(A),P(B)，且若P(AB)=0，則P(A)=P(B)不一定成立，則P(A)+P(B)=1不一定成立， 若P+P=1，則P，故即PB(=PA.B.C.D.所以P(B(=P(AB(+P(A-B(=P(A(+P(A-(?P(B|A-(=+×=，所以P(A|B(==．A12)A.A1,A2是互斥事件B.A1,B2是獨立事件C.P(A1B2(=D.P(B對于BC，P(A1B2(=P(B2|A1(×P(A1(=×=，P(B2(=P(A1(×P(B2|A1(+P(A2(×P(B2|A2(=×+×=，P(A1(P(B2(=×=≠,故A1,B2不是獨立事件，故BC錯誤；對于D，P(B1(=P(A1(×P(B1|A1(+P(A2(×P(B1|A2【答案】/0.4n設(shè)P(An(=pn，依題可知P(Bn(=1-pn，則P(An+1(=P(AnAn+1(+P(BnAn+1(=P(An(P(An+1|An(+P(Bn(P(An+1|Bn(，即Pn+1=0.6Pn+0.2(1-Pn(=0.4P構(gòu)造等比數(shù)列{pn+λ{，設(shè)pn+1+λ=(pn+λ(，解得λ=-，則pn+1-=pn-，又p1=即pn-=×n-1，pn=×n-1+．則第n次聽寫的人是甲的概率為pn=×n-1+.第n次聽寫的人是乙的概率為1-pn=-×n-1.故答案為：;-×n-1 則P(AP(AP(A，所以P(B(=P(B|A0(P(A0(+P(B|A1(P(A1(+P(B|A2(P(A2(根據(jù)題意，有P(B1(=0.4,P(B2(=0.3,P根據(jù)全概率公式，得P(A)=P(A∣B1(P(B1(+P(A∣B2(P(B2(+P(A∣B3(P(B3((1)1.4%；(2)2.(1)丙區(qū)域(1)應(yīng)首先搜索丙區(qū)域.所以P(A)=P(B1(P(AB1(+P(B2(P(A∣B2(+P(B3(P(A∣B