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probability question

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by umaa » Tue Aug 05, 2008 1:00 am
In a particular lottery game, 6 numbers are chosen out of 4 even numbers and 4 odd numbers. What is the probability that the four numbers chosen will have an equal number of even and odd numbers?

A) 1/35
B) 2/7
C) 1/2
D) 4/7
E) 3/4
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by pepeprepa » Tue Aug 05, 2008 1:49 am
Number of total possibilities: 8C6

Number of possibilities you want: (4C3)x(4C3)

Probability is the number of possibilities you want divided by the total number of possibilities:
( (4C3)x(4C3) ) / ( 8C6 ) = 4*4 / 28 = 4/7

That's D
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by parallel_chase » Tue Aug 05, 2008 2:02 am
I agree with you pepeprepa but we need to find out the probability of 4 numbers having equal odd and even integers.

Therefore, after selecting 6, these 4 numbers could be starting 4, middle 4 or ending 4.

Let me know what you think.
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by pepeprepa » Tue Aug 05, 2008 2:08 am
The question does not take into account "order" and factorials neither. For me that is a non-order exercice... I have to work back combinatory to deeply understand but I think it is ok for this question. If you have clues, tell me more.
For me, this one is in the same spirit as forming a group of 4 with 3 boys and 3 girls and having 2 boys and 2 girls. When you make a group like that, order does not matter, you just choose people.
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