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anonymous

  • 4 years ago

Three horses A, B and C are in a race, probability that A wins is twice the probability of that the horse B wins, and probability that B wins is twice the probability of that the C wins. What is the probability that either A or B wins. given, set A={children of both sexes}, B={at most one boy}, if a family have two children then show that the sets are dependent events

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  1. anonymous
    • 4 years ago
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    let x be the probability of C winning. P(A)+P(B)+P(C)=1 4x+2x+x=1 7x=1 x=1/7 P(A)=4/7 and the P(B)=2/7 P(A or B)=4/7+2/7 =6/7

  2. perl
    • 4 years ago
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    how do you go from horses to boys?

  3. anonymous
    • 4 years ago
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    there are two questions, I believe

  4. perl
    • 4 years ago
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    ok , whats the second question?

  5. anonymous
    • 4 years ago
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    given, set A={children of both sexes}, B={at most one boy}, if a family have two children then show that the sets are dependent events

  6. perl
    • 4 years ago
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    so what is A? A = { boy 1 , boy 2 , female 1, female 2} . etc

  7. anonymous
    • 4 years ago
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    hey guyz plz send me finally ans of above this question plz its urgent! thankx

  8. perl
    • 4 years ago
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    see above , Zed

  9. perl
    • 4 years ago
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    so if a family has two kids, then A = { MM , MF, FM, FF} for the possibilities and B = { FM, MF}

  10. anonymous
    • 4 years ago
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    @perl if A is a set of children of both sexes then would it be A={MF,FM} and if B is a set with at most one boy then it would be B={FF,MF,FM}?

  11. anonymous
    • 4 years ago
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    When events are dependent, each possible outcome is related to the other. Given two events A and B, the probability of obtaining both A and B is the product of the probability of obtaining one of the events times the conditional probability of obtaining the other event, given the first event has occurred. P(A and B) = P(A) . P(B|A) This rule says that for both of the two events to occur, the first one must occur ( (P(A) ) and then, given the first event has occurred, the second event occurs ( (P(B|A) ). http://library.thinkquest.org/11506/prules.html

  12. perl
    • 4 years ago
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    one second

  13. perl
    • 4 years ago
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    ok how would we apply the dependence here , in what way

  14. perl
    • 4 years ago
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    ok i redid this, a little confusing. the universal set is U = { MM, MF, FM, FF} now, A = {MF, FM} and B = { MF, FM, FF} P(A) = 2/4 , P(B) = 3/4

  15. anonymous
    • 4 years ago
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    @perl, that looks good.

  16. perl
    • 4 years ago
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    now we compare P(B|A) , and P(B)

  17. anonymous
    • 4 years ago
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    I got P(B|A)=P(B)=3/4

  18. perl
    • 4 years ago
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    I got P(B|A) = 1

  19. anonymous
    • 4 years ago
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    \[P(B|A)=\frac{P(A and B)}{P(A)}\]\[=\frac{2/4*3/4}{2/4}\]\[=\frac{3}{4}\]

  20. perl
    • 4 years ago
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    P(A and B ) only equals 2/4*3/4 if A and B are independent

  21. perl
    • 4 years ago
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    look at the set A intersect B = { MF, FM}

  22. perl
    • 4 years ago
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    P ( A intersect B ) = P ( A) = 2/4 so we have P(B|A) = P(A&B) / P(A) = (2/4 ) / (2/4) unless im making some error

  23. perl
    • 4 years ago
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    agreed?

  24. perl
    • 4 years ago
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    sorry making you go around in circles , hehe

  25. anonymous
    • 4 years ago
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    no worries :)

  26. perl
    • 4 years ago
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    i disagree what you wrote P(B|A) = 3/4 , i got P(B|A) =1

  27. anonymous
    • 4 years ago
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    ahhh I see, my probability is abit rusty.

  28. perl
    • 4 years ago
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    because Universal set = { MM, MF, FM, FF} now, A = {MF, FM} and B = { MF, FM, FF} P(A) = 2/4 , P(B) = 3/4

  29. perl
    • 4 years ago
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    P ( A intersect B ) = P ( A) = 2/4 so we have P(B|A) = P(A&B) / P(A) = (2/4 ) / (2/4)

  30. perl
    • 4 years ago
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    and the question says , show that they are dependent events. so that sort of says something is wrong

  31. perl
    • 4 years ago
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    and P(A|B) = 2/3 i believe

  32. anonymous
    • 4 years ago
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    what do we have to do to show that they are dependent?

  33. perl
    • 4 years ago
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    if P(B|A) = P(B) then they are independent. but since they are not, they are dependent

  34. perl
    • 4 years ago
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    P(A) = 1/2, but P(A|B) = 2/3 (knowing B increases the probability of A) . so they are dependent

  35. perl
    • 4 years ago
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    in your argument you assumed that P(A&B) = P(A)*P(B), but this is only true if you know they are independent

  36. perl
    • 4 years ago
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    and thats what we are attempting to show. so I did it by looking at the sets themselves

  37. anonymous
    • 4 years ago
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    I get it now (P(A|B)=2/3. You're right. I didn't know that they had to be independent for that rule. Thanks :)

  38. perl
    • 4 years ago
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    in fact, the rule is , in general P(A&B) = P(A) * P(B|A)

  39. perl
    • 4 years ago
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    no problem :) and for your horse problem above we can assume that the events are mutually exclusive since you cant have more than one winner

  40. anonymous
    • 4 years ago
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    I hope djsheikh will read this through to see the answers

  41. perl
    • 4 years ago
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    the general addition P(A or B) = P(A) + P(B) - P ( A & B)

  42. perl
    • 4 years ago
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    it would be nice, i see a lot of people run away after they ask their question :(

  43. anonymous
    • 4 years ago
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    So for the horse question the answer should be 2/7?

  44. perl
    • 4 years ago
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    6/7 is right

  45. perl
    • 4 years ago
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    you had it right above :)

  46. anonymous
    • 4 years ago
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    ahh okay is that general addition rule for dependent variables?

  47. perl
    • 4 years ago
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    yes, for two events. gets more complicated for three events

  48. perl
    • 4 years ago
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    but since P(A&B) = 0, since you cant have two winners

  49. perl
    • 4 years ago
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    P( A wins and B wins ) is zero, cant both win

  50. anonymous
    • 4 years ago
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    of course. thanks for clearing it up

  51. perl
    • 4 years ago
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    your welcome, but you did it get started , :))

  52. anonymous
    • 4 years ago
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    given, set A={children of both sexes}, B={at most one boy}, if a family have two children then show that the sets are dependent events ???

  53. anonymous
    • 4 years ago
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    sets are dependent a priori, no probability needed

  54. perl
    • 4 years ago
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    what do you mean ?

  55. perl
    • 4 years ago
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    well i showed with probability, but im not sure what dependent sets are.

  56. perl
    • 4 years ago
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    given, set A={children of both sexes}, B={at most one boy}, if a family have two children then show that the sets are dependent events

  57. perl
    • 4 years ago
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    hi djsheik

  58. anonymous
    • 4 years ago
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    hi

  59. perl
    • 4 years ago
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    so you got the answer ?

  60. perl
    • 4 years ago
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    the universal set for a family of 2 kids is U = { MM, MF, FM, FF} A = {MF, FM} and B = { MF, FM, FF} so P(A) = 2/4 , P(B) = 3/4

  61. anonymous
    • 4 years ago
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    hmm thank you

  62. perl
    • 4 years ago
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    read it and weep buddy, this is the work you should have done

  63. perl
    • 4 years ago
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    so why are A and B dependent?

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