import{E as j,c as _,bE as Q,aj as W,r as P,q as b,C as c,y as o,f as e,o as q,ci as D,l as F,Z as S,W as T,ar as O}from"./index-XCg2QAX4.js";import{b as $}from"./style-MaFG70fX.js";import{d as v}from"./deprecatedFractions-bUE3SVly.js";const G="Conditional probabilities",H="25/10/2021",N="24/10/2021",J="9ccfd",M="1P10";function R(){j.call(this),this.consigne="",this.nbQuestions=1,this.tailleDiaporama=3,this.spacing=_.isHtml?2:1,this.spacingCorr=_.isHtml?3:1,this.video="",this.sup=!1,this.sup2=3;function t(a,C){return C?S(T(a,3)).toLatex().replace("frac","dfrac"):O(T(a,3)).toString().replace(". ","{,}")}this.nouvelleVersion=function(){this.listeQuestions=[],this.listeCorrections=[],this.autoCorrection=[];let a,C;this.sup2=Q(1,3,this.sup2,3),this.sup2===3?C=W(["subjectE3C1","subjectE3C2"],this.nbQuestions):C=W([`subjectE3C${this.sup2}`],this.nbQuestions);for(let A=0,i,h,p,l,u,n,f,m,E,g,k,V,w,y,d,r,s,x=0;A<this.nbQuestions&&x<50;){switch(a=[],C[A]){case"subjectE3C1":i=P(30,70),u=P(30,70),n=P(20,i-5),w=b(.6,2.3),f=b(5,5),m=b(5,1),E=b(9,7.5),g=b(9,4),k=b(9,3),V=b(9,0),a.push(c(w,f,"blue")),a.push(c(w,m,"blue")),a.push(c(f,E,"blue")),a.push(c(f,g,"blue")),a.push(c(m,k,"blue")),a.push(c(m,V,"blue")),a.push(o("A",5,5.2,"black",20,12,"white",10)),a.push(o("\\bar A",5,1.3,"black",20,12,"white",10)),a.push(o("\\Omega",0,2.3,"black",20,12,"white",10)),a.push(o(t(i/100,this.sup),2.5,4.1,"black",20,20,"white",6)),a.push(o(t(1-i/100,this.sup),2.5,2.1,"black",20,20,"white",6)),a.push(o(t(1-u/100,this.sup),6.8,.9,"black",20,20,"white",6)),a.push(o(t(u/100,this.sup),6.8,2.7,"black",20,20,"white",6)),a.push(o(`P(A\\cap V)=${t(n/100,this.sup)}`,10.5,7.8,"red",20,20,"white",10)),a.push(o("V",9,7.7,"black",20,12,"white",10)),a.push(o("\\bar V",9,4.3,"black",20,12,"white",10)),a.push(o("V",9,3.1,"black",20,12,"white",10)),a.push(o("\\bar V",9,.2,"black",20,12,"white",10)),r="A travel agency offers two weekend packages to go to London from Paris. ",r+="<br> Customers choose their means of transport: train or plane. ",r+=`<br> In addition, if they wish, they can complete their formula with the “${e(1)}guided tours${e(1)}” option.`,r+=`<br> A study produced the following ${e(1)} data:`,r+=`<br> $\\bullet ${e(3)} ${i}${e()}\\%$ of customers opt for plane;`,r+=`<br> $\\bullet ${e(3)}$Among customers who have chosen the train, $${u}${e()}\\%$ also choose the “${e(1)}guided tours${e(1)}” option.`,r+=`<br> $\\bullet ${e(3)} ${n}${e()}\\%$ of customers chose both the plane and the “${e(1)}guided tours${e(1)}” option.<br>`,r+="<br> We randomly interview a client of the agency who has subscribed to a weekend package in London. ",r+=`<br> We consider the following events${e(1)}:`,r+="<br> $\\bullet~~$ $A$: the customer has chosen the plane. ",r+=`<br> $\\bullet~~$ $V$: the customer has chosen the “${e(1)}guided tours${e(1)}” option.<br>`,r+=`<br> ${$("1. ")} Determine $P_A(V)$.`,r+=`<br> ${$("2. ")} Demonstrate that the probability that the customer interviewed chose the “${e(1)}guided tours${e(1)}” option is approximately equal to $${t(n/100+(1-i/100)*u/100,!1)}$.`,r+=`<br> ${$("3. ")} Calculate the probability that the customer interviewed took the plane knowing that he did not choose the “${e(1)}guided tours${e(1)}” option. Round the result to the hundredths.`,r+=`<br> ${$("4. ")} Two customers are randomly and independently interviewed.`,r+=`<br> What is the probability that neither of them takes the “${e(1)}guided tours${e(1)}”${e(1)} option?`,r+="We will give the results in the form of approximate values   to the nearest $10^{-3}$. ",s=`${$("1. ")} From the statement, we deduce that${e()}:`,s+=`<br> $P(A)=${t(i/100,this.sup)}$`,s+=`<br> $P_{\\bar{A}}(V)=${t(u/100,this.sup)}$`,s+=`<br> $P(A \\cap V)=${t(n/100,this.sup)}$`,s+=`<br>We can then construct this probability tree${e(1)}: <br>`,s+=q({xmin:-5,ymin:-1,xmax:18,ymax:10},a),s+=`<br>We therefore have $P_{A}(V)=\\dfrac{P(A \\cap V)}{P(A)}=\\dfrac{${t(n/100,this.sup)}}{${t(i/100,this.sup)}}=${v(n,i)} $.`,s+=`<br><br>${$("2. ")} As $A$ and $\\bar A$ form a partition of the universe, we can apply the law of total probabilities${e()}:`,s+=" <br>$P(V)=P(A \\cap V)+P(\\bar{A} \\cap V). $",s+=`<br>Or $P(\\bar{A} \\cap V)=P(\\bar{A}) \\times P_{\\bar{A}}(V)=(1-${t(i/100,this.sup)}) \\times ${t(u/100,this.sup)}=${t((1-i/100)*u/100,this.sup)}$.`,s+=`<br>So $P(V)=${t(n/100,this.sup)}+${t((1-i/100)*u/100,this.sup)}=${t(n/100+(1-i/100)*u/100,this.sup)}$.`,s+=`<br><br>${$("3. ")} We have $P_{\\bar{V}}(A)=\\dfrac{P(\\bar{V} \\cap A)}{P(\\bar{V} )}=\\dfrac{P(A \\cap \\bar{V})}{P(\\bar{V})}=\\dfrac{P(A) \\times P_A(\\bar{ V})}{P(\\bar{V})}$.`,s+=`<br>Now, according to the previous question${e(1)}:${e(1)}$P(\\bar{V})=1-P(V)=1-${t(n/100+(1-i/100)*u/100,this.sup)}=${t(1-(n/100+(1-i/100)*u/100),this.sup)}$`,s+=`<br>and according to question $1: P_{A}(\\bar{V})=1-P_{A}(V)=1-${v(n,i)}=${v(i-n,i)}$.`,y=(i-n)/i,d=1-(n/100+(1-i/100)*u/100),s+=`<br>So $P_{\\bar{V}}(A)=\\dfrac{${t(i/100,this.sup)} \\times ${v(i-n,i)}}{${t(d,this.sup)}} ${D(i/100*y/d,3)}${t(i/100*y/d,!1)}$.`,s+=`<br><br>${$("4. ")} We saw that $P(\\bar{V})=1-${t(d,this.sup)}=${t(1-d,this.sup)}$.`,s+="<br>As the two events are independent, by calling them $\\bar {V_1}$ and $\\bar{V_2}$, we have: $P(\\bar{V_1}\\cap\\bar{ V_2})=P(\\bar{V_1})\\times P(\\bar{V_2})$",s+=`<br>The probability sought is therefore equal to $P(\\bar{V_1}\\cap\\bar{V_2})=${t(1-d,this.sup)} \\times ${t(1-d,this.sup)}\\approx${t((1-d)**2,!1)}$.`;break;case"subjectE3C2":h=P(30,70),p=P(20,95-h),l=P(20,95-h),w=b(.6,2.3),f=b(5,5),m=b(5,1),E=b(9,7.5),g=b(9,4),k=b(9,3),V=b(9,0),a.push(c(w,f,"blue")),a.push(c(w,m,"blue")),a.push(c(f,E,"blue")),a.push(c(f,g,"blue")),a.push(c(m,k,"blue")),a.push(c(m,V,"blue")),a.push(o("C",5,5.2,"black",20,12,"white",10)),a.push(o("\\bar C",5,1.3,"black",20,12,"white",10)),a.push(o("\\Omega",0,2.3,"black",20,12,"white",10)),a.push(o(t(h/100,this.sup),2.5,4.1,"black",20,20,"white",6)),a.push(o(t(1-h/100,this.sup),2.5,2.1,"black",20,20,"white",6)),a.push(o(t(1-p/100,this.sup),6.8,.9,"black",20,20,"white",6)),a.push(o(t(p/100,this.sup),6.8,2.7,"black",20,20,"white",6)),a.push(o(`P(C\\cap E)=${t(l/100,this.sup)}`,10.5,7.8,"red",20,20,"",10)),a.push(o("E",9,7.7,"black",20,12,"white",10)),a.push(o("\\bar E",9,4.3,"black",20,12,"white",10)),a.push(o("E",9,3.1,"black",20,12,"white",10)),a.push(o("\\bar E",9,.2,"black",20,12,"white",10)),r=`A chain of hair salons offers its customers who come for a haircut two additional services that can be combined with ${e()}:`,r+=`<br>$\\bullet$ A natural plant-based coloring called “${e(1)}couleur-soin${e(1)}”, `,r+=`<br>$\\bullet$ Blonde highlights to give relief to the hair, called “${e(1)} sunburn effect${e(1)}”.`,r+=`<br><br> It appears that: <br>$\\diamond ${e(3)} ${h}${e()}\\%$ customers request a “${e(1)}color-care${e(1)}”.`,r+=`<br>$\\diamond ${e(3)}$ Among those who do not want a “${e(1)}color-care${e(1)}”, $${p}${e()}\\%$ of customers ask for a “${e(1)} sunburn effect${e(1)}”.`,r+=`<br>$\\diamond ${e(3)}$ Furthermore, $${l}${e()}\\%$ customers request a “${e(1)}color-care${e(1)}” and a “${e(1)}sunburn effect${e(1)}”.`,r+="<br>We interview a customer at random. ",r+=`<br>We will note $C$ the event: “${e(1)}The customer wants a “${e(1)}color-care${e(1)}”.`,r+=`<br>We will note $E$ the event: “${e(1)}The client wants a “${e(1)} sunburn effect${e(1)}”.<br>`,r+=$("1. ")+" Give the values   of $P(C)$, $P( C \\cap E)$ and $P_{\\bar{C}}(E)$.<br>",r+=$("2. ")+` Calculate the probability that the client wants neither a “${e(1)}color-care${e(1)}” nor a “${e(1)}sunburn effect${e(1)}”.<br>`,r+=$("3. ")+` Calculate the probability that a client chooses the “${e(1)} sunburn effect${e(1)}” knowing that they have taken a “${e(1)}color-care${e(1)}”.<br>`,r+=$("4. ")+` Show that the probability of the event $E$ is equal to $${t(l/100+(1-h/100)*p/100,!1)}$ (within $10^{-3}$).<br>`,r+=$("5. ")+" Are the events $C$ and $E$ independent?<br>",r+="We will give the results in the form of approximate values   to the nearest $10^{-3}$. ",s=`${$("1. ")} According to the statement, we have:<br>$\\bullet~~P(C)=${t(h/100,this.sup)}$`,s+=`<br>$\\bullet~~P(C \\cap E)=${t(l/100,this.sup)}$`,s+=`<br>$\\bullet~~P_{\\bar C}(E)=${t(p/100,this.sup)}$`,s+=`<br>Which allows us to construct this probability tree${e()}:${e()}`,s+=q({xmin:-5,ymin:-1,xmax:18,ymax:10},a),s+=`<br>${$("2. ")} The event: the client does not want a “${e(1)}color-care${e(1)}” nor a “${e(1)}sunburn effect${e(1)}” corresponds to $\\bar{C} \\cap \\bar{E}$.`,s+=`<br>We have $P(\\bar{C} \\cap \\bar{E})=P(\\bar{C}) \\times P_{\\bar{C}}(\\bar {E})=P(\\bar{C}) \\times (1-P_{\\bar{C}}(E))=${t(1-h/100,!1)} \\times ${t(1-p/100,!1)}\\approx ${t((1-h/100)*(1-p/100),!1)}$.`,s+=`<br>${$("3. ")} The probability that a customer chooses the “${e(1)} sunburn effect${e(1)}” knowing that he has taken a “${e(1)}color-care${e(1)}” is $P_{C}(E)$.`,s+=`<br>We then have according to the weighted tree${e()}:`,s+=`<br>$P(C) \\times P_{C}(E)=${t(h/100,!1)} \\times P_{C}(E)=${t(l/100,!1)}$.`,s+=`<br>We deduce that $P_{C}(E)=\\dfrac{${t(l/100,!1)}}{${t(h/100,!1)}}\\approx${t(l/h,!1)}$.`,s+=`<br>${$("4. ")} We are looking for $P(E)$ which is a total probability.`,s+=`<br>As $C$ and $\\bar C$ form a partition of the universe, we can apply the law of total probabilities${e()}:`,s+="<br>$P(E)=P(E \\cap C)+P(E \\cap \\bar{C} )$",s+=`<br>$P(E)=${t(l/100,!1)}+${t(1-h/100,!1)}\\times ${t(p/100,!1)}$`,s+=`<br>$P(E)\\approx${t(l/100+(1-h/100)*p/100,!1)}$`,s+=`<br>${$("5. ")} To know if the events $C$ and $E$ are independent, we calculate separately `,s+="$P(C \\cap E)$ and $P(C) \\times P(E)$, to test if they are equal. ",s+=`<br>We have $P(C \\cap E)=${t(l/100,!1)}$`,s+=`and $P(C) \\times P(E)\\approx${t(h/100*(l/100+(1-h/100)*p/100),!1)}$.`,s+="<br>We deduce that the events $C$ and $E$ are not independent. ";break}this.questionJamaisPosee(A,r)&&(this.listeQuestions.push(r),this.listeCorrections.push(s),A++),x++}F(this)},this.besoinFormulaireCaseACocher=["Fractional probabilities",!1],this.besoinFormulaire2Numerique=["Choice of exercises:",3,`1: Subject 1 from E3C
2: Subject 2 from E3C
3: Combination`]}export{N as dateDeModifImportante,H as dateDePublication,R as default,M as ref,G as titre,J as uuid};
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