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Another interesting one from the Hakob's set: What is the

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Another interesting one from the Hakob's set: What is the [#permalink] New post 19 Mar 2003, 09:14
Another interesting one from the Hakob's set:

What is the probability of bearing two boys from four born babies?

A) 1/5
B) 1/8
C) 1/2
D) 3/8
E) 6/11
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 [#permalink] New post 19 Mar 2003, 09:31
isn't it necessary to know how many girls and boys there are in the group of four babies :?:
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 [#permalink] New post 19 Mar 2003, 10:07
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commander wrote:
isn't it necessary to know how many girls and boys there are in the group of four babies :?:



not really, you just start out assuming that you can have only boys and girls; not neutrals :)

Then I think you will need to calculate how many ways you can get 2 boys out of a group of 4, which is C(2,4) = 6

Then probability kicks in....

I am not good in these, but I will try, I got 3/8 :computer
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 [#permalink] New post 19 Mar 2003, 10:22
pleas explain? :help
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 [#permalink] New post 19 Mar 2003, 10:26
okay I got it: It's a combined problem:

chance of someone being a girl or boy is 1/2.

taking 2 out of four can be done in like you said in: 2C4=6 ways

so the probability is: 6x1/2x1/2x1/2x1/2= 3/8

Now how can one remember this rule?
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 [#permalink] New post 19 Mar 2003, 10:32
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commander wrote:
pleas explain? :help



Probability is not my strong point, but I think I got this one :wink: or at least I can explain the logic behind it:
whatever your probability will be to have 2 out of 4, it needs to be multiplied by 6 since you have 6 ways to pick them.

So, the answer basically depends on the probability.

I got the result by 1/2*1/2*1/2*1/2

I assumed you get 1/2 probability to have a boy/girl at any point in time.
So, I got 1/2 for boy, 1/2 for another boy, 1/2 for girl, and 1/2 for girl, (you need to get exactly 2 boys and 2 girls). So, it is 1/16

6*1/16=3/8


--=-
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 [#permalink] New post 19 Mar 2003, 10:33
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commander wrote:
okay I got it: It's a combined problem:

chance of someone being a girl or boy is 1/2.

taking 2 out of four can be done in like you said in: 2C4=6 ways

so the probability is: 6x1/2x1/2x1/2x1/2= 3/8

Now how can one remember this rule?



You were faster than I :)
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 [#permalink] New post 19 Mar 2003, 22:09
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stolyar wrote:
3/8 is the answer



:cooldude
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 [#permalink] New post 10 Sep 2003, 13:11
Isn't there a formula for this type of problem:

nCk * p^k * (1-p)^(n-k)

it works out to 3/8 also

Htown
  [#permalink] 10 Sep 2003, 13:11
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Another interesting one from the Hakob's set: What is the

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