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

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SVP
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Another interesting one from the Hakob's set: What is the [#permalink]  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
Intern
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isn't it necessary to know how many girls and boys there are in the group of four babies
Founder
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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
Intern
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pleas explain? :help
Intern
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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?
Founder
Affiliations: AS - Gold, UA - Silver, HH-Diamond
Joined: 04 Dec 2002
Posts: 12734
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
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Kudos [?]: 12300 [0], given: 3828

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

--=-
Founder
Affiliations: AS - Gold, UA - Silver, HH-Diamond
Joined: 04 Dec 2002
Posts: 12734
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
WE: Information Technology (Hospitality and Tourism)
Followers: 2601

Kudos [?]: 12300 [0], given: 3828

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
Founder
Affiliations: AS - Gold, UA - Silver, HH-Diamond
Joined: 04 Dec 2002
Posts: 12734
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
WE: Information Technology (Hospitality and Tourism)
Followers: 2601

Kudos [?]: 12300 [0], given: 3828

Expert's post
stolyar wrote:

dude
Manager
Joined: 24 Jun 2003
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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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