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QUESTION 1 (NOV.24TH)

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QUESTION 1 (NOV.24TH)

by bacali » Mon Nov 24, 2008 11:45 am
Q5:

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?
A. 1/10
B. 4/9
C. 1/2
D. 3/5
E. 2/3


OA: D
Last edited by bacali on Tue Nov 25, 2008 7:38 am, edited 1 time in total.
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by dmateer25 » Mon Nov 24, 2008 3:35 pm
Ways to choose 2 girls 3C2 = 3
Ways to choose 2 boys 3C2 = 3

Total number of ways to select 4 children = 6C4 = 15

(3*3)/15
=9/15
=3/5
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by gmataug08 » Mon Nov 24, 2008 6:59 pm
the possibility of selecting 4 members has three options

3G+ 1B, 2G+2B , 3B+1G

to find the probabilty of 2G+2B = 1 - [(3G+1B) + (3B+1G)]

3G+1B = 3/6 x 2/5 x 1/4 x 3/3 = 1/20

3B+1G = 1/20

hence, 2G+2B = 1- (2/20) => 1- 1/10 => 9/10
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by cramya » Mon Nov 24, 2008 7:02 pm
One more vote for 3/5 similar to dmatter's approach!
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by jymloke » Mon Nov 24, 2008 8:14 pm
Ditto with 3/5.

(3C2 * 3C2 ) / (6C4)
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by logitech » Mon Nov 24, 2008 9:44 pm
Hey the answer supposed to be 53 :lol:
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by 4meonly » Tue Nov 25, 2008 5:42 am
logitech wrote:Hey the answer supposed to be 53 :lol:
nice probability 53 8) :-)
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by bacali » Tue Nov 25, 2008 7:39 am
Changed it. I copied a lot over and it really messed up in the process.
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