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Pulsified333

  • one year ago

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]=

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  1. Pulsified333
    • one year ago
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    okay

  2. anonymous
    • one year ago
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    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

  3. Pulsified333
    • one year ago
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    is that for (F'∩G)

  4. anonymous
    • one year ago
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    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

  5. Pulsified333
    • one year ago
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    (1) Pr[E′∪F]= .6? (2) Pr[F′∩G]= .2? (3) Pr[E∩G]= .2

  6. Pulsified333
    • one year ago
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    @amirreza1870 ?

  7. anonymous
    • one year ago
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    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)

  8. Pulsified333
    • one year ago
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    ok

  9. anonymous
    • one year ago
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    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)

  10. Pulsified333
    • one year ago
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    how would you find #2 then?

  11. anonymous
    • one year ago
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    P(F′∩G)=P(F')*P(G)

  12. Pulsified333
    • one year ago
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    did you make a venn diagram?

  13. Pulsified333
    • one year ago
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    .78 is wrong

  14. anonymous
    • one year ago
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    when two events are independent P(A∩B)=P(A)*P(B)

  15. anonymous
    • one year ago
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    (1) Pr[E′∪F]= 0.78 (2) Pr[F′∩G]= 0.2475 (3) Pr[E∩G]= 0.2

  16. Pulsified333
    • one year ago
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    1 isnt right

  17. anonymous
    • one year ago
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    every human can make mistakes.can you explain why the first one isn't correct?

  18. Pulsified333
    • one year ago
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    is it because of P(E')*P(F)

  19. anonymous
    • one year ago
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    but it's Pr[E′∪F]

  20. Pulsified333
    • one year ago
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    where did .27 come from?

  21. anonymous
    • one year ago
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    p(E'∩F)=P(E')*P(F)=(1-P(E))*P(F)=(1-0.4)*0.45=0.6*0.45=0.27

  22. Pulsified333
    • one year ago
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    oh

  23. Pulsified333
    • one year ago
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    but then I don't see why its the wrong answer

  24. BAdhi
    • one year ago
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    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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