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by ketkoag » Thu May 07, 2009 5:53 am
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the
probability that equal numbers of boys and girls will be selected?

my ans: [spoiler]3/10[/spoiler] but OA: [spoiler]3/5[/spoiler] how??
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by dmateer25 » Thu May 07, 2009 5:57 am
Total ways to select 4 children from 6. 6C4 = 15


Total ways to select 2 boys from 3 = 3C2 = 3

Total ways to select 2 girls from 3= 3C2 = 3


(3C2 * 3C2)/6C4 = 9/15 = 3/5
Last edited by dmateer25 on Thu May 07, 2009 10:46 am, edited 1 time in total.
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multiply

by madhukumar_v » Thu May 07, 2009 7:38 am
dmateer...

why did you multiply 3*3 instead of adding...

I am always confused when to add and when to multiply. Can you please explain it for me.

Thanks
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by hk » Thu May 07, 2009 8:16 am
madhu,

Always remember in probability or set theory,

AND = multiply
OR = Add

Here the question asks for equal number of boys and girls.. So there MUST be 2 Boys AND 2 girls. So you multiple their individual chances...

To make things clearer, lets suppose the question asks to pick 2 people from this group and there is no restriction on the gender.

Then the possibilities are as follows:

(Boy AND Boy) OR (Girl AND Girl) OR (Boy AND Girl)

Now when you solve remember to multiply the values where you see AND and add the values where you see OR..

Hope this helped.
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by ketkoag » Thu May 07, 2009 9:37 am
dmateer25 wrote:Total ways to select 4 children from 6. 6C4 = 15


Total ways to select 2 boys from 3 = 3C2 = 3

Total ways to select 2 girls from 3= 3C2 = 3


(3C3 * 3C3)/6C4 = 9/15 = 3/5
OMG, i did a silly mistake while solving it....
thanks for ur reply..
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Thank you

by madhukumar_v » Thu May 07, 2009 9:59 am
Thanks Hari...Good Luck with your GMAT.

M
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Thank you

by madhukumar_v » Thu May 07, 2009 10:00 am
Thanks Hari...Good Luck with your GMAT.

M
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