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

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- 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

how do you go from horses to boys?

- anonymous

there are two questions, I believe

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## More answers

- perl

ok , whats the second question?

- 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

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

- anonymous

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

- perl

see above , Zed

- perl

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

- 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

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

one second

- perl

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

- 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

@perl, that looks good.

- perl

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

- anonymous

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

- perl

I got P(B|A) = 1

- anonymous

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

- perl

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

- perl

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

- 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

agreed?

- perl

sorry making you go around in circles , hehe

- anonymous

no worries :)

- perl

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

- anonymous

ahhh I see, my probability is abit rusty.

- 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

P ( A intersect B ) = P ( A) = 2/4 so we have P(B|A) = P(A&B) / P(A) = (2/4 ) / (2/4)

- perl

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

- perl

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

- anonymous

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

- perl

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

- perl

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

- 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

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

- 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

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

- 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

I hope djsheikh will read this through to see the answers

- perl

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

- perl

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

- anonymous

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

- perl

6/7 is right

- perl

you had it right above :)

- anonymous

ahh okay is that general addition rule for dependent variables?

- perl

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

- perl

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

- perl

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

- anonymous

of course. thanks for clearing it up

- perl

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

- 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

sets are dependent a priori, no probability needed

- perl

what do you mean ?

- perl

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

- perl

- perl

hi djsheik

- anonymous

hi

- perl

so you got the answer ?

- 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

hmm thank you

- perl

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

- perl

so why are A and B dependent?

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