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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]

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19 Mar 2003, 10:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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19 Mar 2003, 10:31
isn't it necessary to know how many girls and boys there are in the group of four babies
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19 Mar 2003, 11:07
Expert's post
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
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19 Mar 2003, 11:22
pleas explain? :help
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19 Mar 2003, 11:26
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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19 Mar 2003, 11:32
Expert's post
commander wrote:
pleas explain? :help

Probability is not my strong point, but I think I got this one 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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Kudos [?]: 19458 [0], given: 4260

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19 Mar 2003, 11:33
Expert's post
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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Kudos [?]: 19458 [0], given: 4260

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19 Mar 2003, 23:09
Expert's post
stolyar wrote:

dude
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10 Sep 2003, 14:11
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
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