Pulsified333
  • Pulsified333
events E, F, and G in a sample space S. Assume that Pr[E]=0.4, Pr[F]=0.45, Pr[G]=0.45, Pr[E∪F]=0.6, Pr[E∪G]=0.65, and Pr[F∪G]=0.7. Find the following probabilities: (1) Pr[E′∪F]= (2) Pr[F′∩G]= (3) Pr[E∩G]=
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Pulsified333
  • Pulsified333
okay
anonymous
  • anonymous
P(E∪G)=p(E)+P(G)-P(E∩G) 0.65=0.4+0.45-P(E∩G) P(E∩G)=0.85-0.65=0.2
Pulsified333
  • Pulsified333
is that for (F'∩G)

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anonymous
  • anonymous
now we make make a list of probabilities: Pr[E]=0.4, Pr[F]=0.45, Pr[G]=0.45, Pr[E∪F]=0.6, Pr[E∪G]=0.65, Pr[F∪G]=0.7. P(E∩G)=0.2 P(F∩G)=0.2 p(E∩G)=0.2
Pulsified333
  • Pulsified333
(1) Pr[E′∪F]= .6? (2) Pr[F′∩G]= .2? (3) Pr[E∩G]= .2
Pulsified333
  • Pulsified333
@amirreza1870 ?
anonymous
  • anonymous
now we start to find P(E′∪F)=P(E′)+P(F)-P(E'∩G) THE third one is true but we are asking for P(E'∩G) not p(E∩G)
Pulsified333
  • Pulsified333
ok
anonymous
  • anonymous
sorry.P(E′∪F)=P(E′)+P(F)-P(E'∩F) .P(E′∪F)=0.6+0.45-P(E')*P(F) .P(E′∪F)=0.6+0.45-0.27=0.78 i used these rules:P(A)=1-P(A') when A and B are independent :P(A∩B)=P(A)*P(B)
Pulsified333
  • Pulsified333
how would you find #2 then?
anonymous
  • anonymous
P(F′∩G)=P(F')*P(G)
Pulsified333
  • Pulsified333
did you make a venn diagram?
Pulsified333
  • Pulsified333
.78 is wrong
anonymous
  • anonymous
when two events are independent P(A∩B)=P(A)*P(B)
anonymous
  • anonymous
(1) Pr[E′∪F]= 0.78 (2) Pr[F′∩G]= 0.2475 (3) Pr[E∩G]= 0.2
Pulsified333
  • Pulsified333
1 isnt right
anonymous
  • anonymous
every human can make mistakes.can you explain why the first one isn't correct?
Pulsified333
  • Pulsified333
is it because of P(E')*P(F)
anonymous
  • anonymous
but it's Pr[E′∪F]
Pulsified333
  • Pulsified333
where did .27 come from?
anonymous
  • anonymous
p(E'∩F)=P(E')*P(F)=(1-P(E))*P(F)=(1-0.4)*0.45=0.6*0.45=0.27
Pulsified333
  • Pulsified333
oh
Pulsified333
  • Pulsified333
but then I don't see why its the wrong answer
BAdhi
  • BAdhi
for the first one, \(P[E\cup F] = P[F]+P[E]-P[E\cap F]\) \( P[E\cup F] -P[F]= P[E]-P[E\cap F]\) \(P[E' \cap F] = 1-(P[E]-P[E\cap F]) =1- P[E\cup F] +P[F]\) Please draw a venn diagram if you cannot clarify

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