## 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

1. anonymous

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

how do you go from horses to boys?

3. anonymous

there are two questions, I believe

4. perl

ok , whats the second question?

5. anonymous

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

so what is A? A = { boy 1 , boy 2 , female 1, female 2} . etc

7. anonymous

hey guyz plz send me finally ans of above this question plz its urgent! thankx

8. perl

see above , Zed

9. perl

so if a family has two kids, then A = { MM , MF, FM, FF} for the possibilities and B = { FM, MF}

10. anonymous

@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

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

one second

13. perl

ok how would we apply the dependence here , in what way

14. perl

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

@perl, that looks good.

16. perl

now we compare P(B|A) , and P(B)

17. anonymous

I got P(B|A)=P(B)=3/4

18. perl

I got P(B|A) = 1

19. anonymous

$P(B|A)=\frac{P(A and B)}{P(A)}$$=\frac{2/4*3/4}{2/4}$$=\frac{3}{4}$

20. perl

P(A and B ) only equals 2/4*3/4 if A and B are independent

21. perl

look at the set A intersect B = { MF, FM}

22. perl

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

agreed?

24. perl

sorry making you go around in circles , hehe

25. anonymous

no worries :)

26. perl

i disagree what you wrote P(B|A) = 3/4 , i got P(B|A) =1

27. anonymous

ahhh I see, my probability is abit rusty.

28. perl

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

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

and the question says , show that they are dependent events. so that sort of says something is wrong

31. perl

and P(A|B) = 2/3 i believe

32. anonymous

what do we have to do to show that they are dependent?

33. perl

if P(B|A) = P(B) then they are independent. but since they are not, they are dependent

34. perl

P(A) = 1/2, but P(A|B) = 2/3 (knowing B increases the probability of A) . so they are dependent

35. perl

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

and thats what we are attempting to show. so I did it by looking at the sets themselves

37. anonymous

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

in fact, the rule is , in general P(A&B) = P(A) * P(B|A)

39. perl

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

41. perl

the general addition P(A or B) = P(A) + P(B) - P ( A & B)

42. perl

it would be nice, i see a lot of people run away after they ask their question :(

43. anonymous

So for the horse question the answer should be 2/7?

44. perl

6/7 is right

45. perl

you had it right above :)

46. anonymous

ahh okay is that general addition rule for dependent variables?

47. perl

yes, for two events. gets more complicated for three events

48. perl

but since P(A&B) = 0, since you cant have two winners

49. perl

P( A wins and B wins ) is zero, cant both win

50. anonymous

of course. thanks for clearing it up

51. perl

your welcome, but you did it get started , :))

52. anonymous

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

sets are dependent a priori, no probability needed

54. perl

what do you mean ?

55. perl

well i showed with probability, but im not sure what dependent sets are.

56. perl

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

hi djsheik

58. anonymous

hi

59. perl

so you got the answer ?

60. perl

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

hmm thank you

62. perl

read it and weep buddy, this is the work you should have done

63. perl

so why are A and B dependent?