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The probability that a family with 6 children has exactly

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Re: The probability that a family with 6 children has exactly  [#permalink]

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New post 09 Sep 2018, 06:44
C as 15 favorable and each position can take 2 either boy or girl so we have 6 position i.e 64

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Re: The probability that a family with 6 children has exactly  [#permalink]

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New post 10 Sep 2018, 01:02
I have a doubt e.g: question i-3 m and 5 women. We need to select 5 people . P(2 are men)

So i do Total outcomes 8C2 Favorable 3c2* 5c3 ans= fav/total.

Here in this question if they have asked me P of exactly 4 boys. shouldn't Favorable Outcomes be equal to --->be 6c4(boys)*6c2(if not boys then girls)

Need guidance on what the difference int he given question and a assumed question i mentioned above.

Thanks in advance

Bunuel wrote:
ashueureka wrote:
I think this way:

For each child we've two options i.e., that child can either be a boy or a girl.Hence we have 2^6 = 64 total outcomes.
Now out of 6 children any 4 of them can boys and this can be done in 6C4 = 15 ways.
Now here I don't think that ordering does matter in this case.

So the probability comes out to be 15/64.

P.S. Please explain me if ordering really matters in this case as my understanding is quite naive in these problems.


This way of solving is also correct. You've already considered all possible arrangements for BBBBGG with 6C4, which is 6!/4!2!. If you look at my solution and at yours you'll see that both wrote the same formulas but with different approach.
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Re: The probability that a family with 6 children has exactly &nbs [#permalink] 10 Sep 2018, 01:02

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