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Probability Question

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Master | Next Rank: 500 Posts
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Re: Probability Question

by Bidisha800 » Sun Feb 15, 2009 2:20 pm
Total possibility = 6C4= 6!/4!2!=15

3 girls and I boy can be selected 1X3 = 3 ways
3 buys and 1 girl can b selected 1 X 3 = 3 ways
2 boys and 3gors can be selected in 9 ways


therefore, probability of selecting equal number of girls and boys = 9/15
=3/5 = 60%
Drill baby drill !

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by dendude » Mon Feb 16, 2009 2:32 pm
peddisetty wrote:Why cant we pick 3 boys 3 girls and 1 boy 1 girl ?
Why is it only two 2 boys and 2 girsl?.

In the question it is asked, equal no of boys and equal no of girls.

Please somebody clarify.

Thanks.
Because, as per the question, 4 children are to be selected.
2 Boys can be chosen in 3C2 = 3 ways
2 Girls can be chosen in 3C2 = 3 ways
And 4 children can be chosen in 6C4 = 15 ways

Using this info, the prob that 2 boys and 2 girls will be chosen is (3*3)/15 = 3/5
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Why?

by ashishsj » Wed Feb 25, 2009 8:34 am
peddisetty wrote:Why cant we pick 3 boys 3 girls and 1 boy 1 girl ?
Why is it only two 2 boys and 2 girsl?.

In the question it is asked, equal no of boys and equal no of girls.

Please somebody clarify.

Thanks.
Because problem states that 4 children have to be selected!
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