anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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
perl
  • perl
how do you go from horses to boys?
anonymous
  • anonymous
there are two questions, I believe

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perl
  • perl
ok , whats the second question?
anonymous
  • 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
perl
  • perl
so what is A? A = { boy 1 , boy 2 , female 1, female 2} . etc
anonymous
  • anonymous
hey guyz plz send me finally ans of above this question plz its urgent! thankx
perl
  • perl
see above , Zed
perl
  • perl
so if a family has two kids, then A = { MM , MF, FM, FF} for the possibilities and B = { FM, MF}
anonymous
  • 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}?
anonymous
  • 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
perl
  • perl
one second
perl
  • perl
ok how would we apply the dependence here , in what way
perl
  • 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
anonymous
  • anonymous
@perl, that looks good.
perl
  • perl
now we compare P(B|A) , and P(B)
anonymous
  • anonymous
I got P(B|A)=P(B)=3/4
perl
  • perl
I got P(B|A) = 1
anonymous
  • anonymous
\[P(B|A)=\frac{P(A and B)}{P(A)}\]\[=\frac{2/4*3/4}{2/4}\]\[=\frac{3}{4}\]
perl
  • perl
P(A and B ) only equals 2/4*3/4 if A and B are independent
perl
  • perl
look at the set A intersect B = { MF, FM}
perl
  • 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
perl
  • perl
agreed?
perl
  • perl
sorry making you go around in circles , hehe
anonymous
  • anonymous
no worries :)
perl
  • perl
i disagree what you wrote P(B|A) = 3/4 , i got P(B|A) =1
anonymous
  • anonymous
ahhh I see, my probability is abit rusty.
perl
  • 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
perl
  • perl
P ( A intersect B ) = P ( A) = 2/4 so we have P(B|A) = P(A&B) / P(A) = (2/4 ) / (2/4)
perl
  • perl
and the question says , show that they are dependent events. so that sort of says something is wrong
perl
  • perl
and P(A|B) = 2/3 i believe
anonymous
  • anonymous
what do we have to do to show that they are dependent?
perl
  • perl
if P(B|A) = P(B) then they are independent. but since they are not, they are dependent
perl
  • perl
P(A) = 1/2, but P(A|B) = 2/3 (knowing B increases the probability of A) . so they are dependent
perl
  • 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
perl
  • perl
and thats what we are attempting to show. so I did it by looking at the sets themselves
anonymous
  • 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 :)
perl
  • perl
in fact, the rule is , in general P(A&B) = P(A) * P(B|A)
perl
  • 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
anonymous
  • anonymous
I hope djsheikh will read this through to see the answers
perl
  • perl
the general addition P(A or B) = P(A) + P(B) - P ( A & B)
perl
  • perl
it would be nice, i see a lot of people run away after they ask their question :(
anonymous
  • anonymous
So for the horse question the answer should be 2/7?
perl
  • perl
6/7 is right
perl
  • perl
you had it right above :)
anonymous
  • anonymous
ahh okay is that general addition rule for dependent variables?
perl
  • perl
yes, for two events. gets more complicated for three events
perl
  • perl
but since P(A&B) = 0, since you cant have two winners
perl
  • perl
P( A wins and B wins ) is zero, cant both win
anonymous
  • anonymous
of course. thanks for clearing it up
perl
  • perl
your welcome, but you did it get started , :))
anonymous
  • 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 ???
anonymous
  • anonymous
sets are dependent a priori, no probability needed
perl
  • perl
what do you mean ?
perl
  • perl
well i showed with probability, but im not sure what dependent sets are.
perl
  • 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
perl
  • perl
hi djsheik
anonymous
  • anonymous
hi
perl
  • perl
so you got the answer ?
perl
  • 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
anonymous
  • anonymous
hmm thank you
perl
  • perl
read it and weep buddy, this is the work you should have done
perl
  • perl
so why are A and B dependent?

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