I am having actually trouble understanding how the probcapacity of at least one girl is 3/4 in this question listed below.

You are watching: A family has five kids. what is the probability that they consist of at least one female?

Given a family members of 2 youngsters (assume boys and girls equally likely, that is, probability 1/2 for each), That at least one is a girl?

My factor for the confusion is that, if at least one is a girl then the probability of having actually a boy-boy does not exist, therefore the sample space becomes

so sudepend the answer need to be 2/3


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Your sample area is not $.$

Given a family members of two kids (assume boys and also girls equally likely, that is, probability 1/2 for each) ...

This provides us (so far) the sample room $.$

... what is the probcapacity ...

You left these words out of the question, but I think they were implied.They intend that we have to now gain prepared to compute the probcapability of some occasion.

... that at least one is a girl?

OK, so the occasion is "at leastern one girl." The complying with facets of your sample room have at leastern one girl: $GG, BG, GB.$ The aspect $BB$ does not.

So the event we are asked to provide the probcapability of is $.$

The sample space is still $.$

We are evidently meant to assume that the births are independent as well as equally most likely to be boys or girls, so the facets of the sample space are equally most likely.Thus an event through $3$ of the $4$ aspects has actually probcapacity $3/4.$


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answered Aug 18 "18 at 16:16
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David KDavid K
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The question does not assume tright here is at least one daughter in the household. The just presumption around children is tright here are two. Hence the prob. room is (BG, GB, BB, GG) and also 3 of the four elementary events meet the requirement.


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answered Aug 18 "18 at 16:15
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CiaPanCiaPan
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It isn"t tough to list the instances. Because each has a probcapability of $frac12,$ tright here are equally most likely cases $GG,BG,GB,BB.$ Out of the 4 instances, $3$ have at leastern one girl, so it is $frac34.$Complementary counting likewise functions. The probability of no girls (all boys) is $left(frac12 ight)^2=frac14.$ The probcapacity of at leastern one girl is $1-frac14=frac34.$Either way, the probability is $frac34.$It appears you are answering the different problem

If tbelow is at least one girl in a household with $2$ youngsters, then what is the probcapacity of just one girl?


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edited Aug 18 "18 at 16:13
answered Aug 18 "18 at 16:11
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Jakid KimJason Kim
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The probcapability of having actually two boys is 1/4, not 0.

eginalign P(child1=boy) imes P(child2=boy) = frac12 imesfrac12\ = frac14.endalign


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answered Aug 18 "18 at 16:50
DiatcheDiatche
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No, equally likely either GB or BG so answer is 1/2


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answered Oct 25 "19 at 15:58
PreethamPreetham
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