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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
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

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let 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

perl
 4 years ago
Best ResponseYou've already chosen the best response.2how do you go from horses to boys?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there are two questions, I believe

perl
 4 years ago
Best ResponseYou've already chosen the best response.2ok , whats the second question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0given, 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

perl
 4 years ago
Best ResponseYou've already chosen the best response.2so what is A? A = { boy 1 , boy 2 , female 1, female 2} . etc

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey guyz plz send me finally ans of above this question plz its urgent! thankx

perl
 4 years ago
Best ResponseYou've already chosen the best response.2so if a family has two kids, then A = { MM , MF, FM, FF} for the possibilities and B = { FM, MF}

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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}?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0When 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(BA) 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(BA) ). http://library.thinkquest.org/11506/prules.html

perl
 4 years ago
Best ResponseYou've already chosen the best response.2ok how would we apply the dependence here , in what way

perl
 4 years ago
Best ResponseYou've already chosen the best response.2ok 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@perl, that looks good.

perl
 4 years ago
Best ResponseYou've already chosen the best response.2now we compare P(BA) , and P(B)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got P(BA)=P(B)=3/4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[P(BA)=\frac{P(A and B)}{P(A)}\]\[=\frac{2/4*3/4}{2/4}\]\[=\frac{3}{4}\]

perl
 4 years ago
Best ResponseYou've already chosen the best response.2P(A and B ) only equals 2/4*3/4 if A and B are independent

perl
 4 years ago
Best ResponseYou've already chosen the best response.2look at the set A intersect B = { MF, FM}

perl
 4 years ago
Best ResponseYou've already chosen the best response.2P ( A intersect B ) = P ( A) = 2/4 so we have P(BA) = P(A&B) / P(A) = (2/4 ) / (2/4) unless im making some error

perl
 4 years ago
Best ResponseYou've already chosen the best response.2sorry making you go around in circles , hehe

perl
 4 years ago
Best ResponseYou've already chosen the best response.2i disagree what you wrote P(BA) = 3/4 , i got P(BA) =1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahhh I see, my probability is abit rusty.

perl
 4 years ago
Best ResponseYou've already chosen the best response.2because Universal set = { MM, MF, FM, FF} now, A = {MF, FM} and B = { MF, FM, FF} P(A) = 2/4 , P(B) = 3/4

perl
 4 years ago
Best ResponseYou've already chosen the best response.2P ( A intersect B ) = P ( A) = 2/4 so we have P(BA) = P(A&B) / P(A) = (2/4 ) / (2/4)

perl
 4 years ago
Best ResponseYou've already chosen the best response.2and the question says , show that they are dependent events. so that sort of says something is wrong

perl
 4 years ago
Best ResponseYou've already chosen the best response.2and P(AB) = 2/3 i believe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what do we have to do to show that they are dependent?

perl
 4 years ago
Best ResponseYou've already chosen the best response.2if P(BA) = P(B) then they are independent. but since they are not, they are dependent

perl
 4 years ago
Best ResponseYou've already chosen the best response.2P(A) = 1/2, but P(AB) = 2/3 (knowing B increases the probability of A) . so they are dependent

perl
 4 years ago
Best ResponseYou've already chosen the best response.2in your argument you assumed that P(A&B) = P(A)*P(B), but this is only true if you know they are independent

perl
 4 years ago
Best ResponseYou've already chosen the best response.2and thats what we are attempting to show. so I did it by looking at the sets themselves

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I get it now (P(AB)=2/3. You're right. I didn't know that they had to be independent for that rule. Thanks :)

perl
 4 years ago
Best ResponseYou've already chosen the best response.2in fact, the rule is , in general P(A&B) = P(A) * P(BA)

perl
 4 years ago
Best ResponseYou've already chosen the best response.2no problem :) and for your horse problem above we can assume that the events are mutually exclusive since you cant have more than one winner

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I hope djsheikh will read this through to see the answers

perl
 4 years ago
Best ResponseYou've already chosen the best response.2the general addition P(A or B) = P(A) + P(B)  P ( A & B)

perl
 4 years ago
Best ResponseYou've already chosen the best response.2it would be nice, i see a lot of people run away after they ask their question :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So for the horse question the answer should be 2/7?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh okay is that general addition rule for dependent variables?

perl
 4 years ago
Best ResponseYou've already chosen the best response.2yes, for two events. gets more complicated for three events

perl
 4 years ago
Best ResponseYou've already chosen the best response.2but since P(A&B) = 0, since you cant have two winners

perl
 4 years ago
Best ResponseYou've already chosen the best response.2P( A wins and B wins ) is zero, cant both win

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0of course. thanks for clearing it up

perl
 4 years ago
Best ResponseYou've already chosen the best response.2your welcome, but you did it get started , :))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0given, 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 ???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sets are dependent a priori, no probability needed

perl
 4 years ago
Best ResponseYou've already chosen the best response.2well i showed with probability, but im not sure what dependent sets are.

perl
 4 years ago
Best ResponseYou've already chosen the best response.2given, 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

perl
 4 years ago
Best ResponseYou've already chosen the best response.2the 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

perl
 4 years ago
Best ResponseYou've already chosen the best response.2read it and weep buddy, this is the work you should have done

perl
 4 years ago
Best ResponseYou've already chosen the best response.2so why are A and B dependent?
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