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Combination & Probability 1

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by karthikpandian19 » Thu Dec 22, 2011 9:48 pm
In a particular lottery game, 6 numbers are chosen out of 4 even numbers and 4 odd numbers. What is the probability that the six 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

Can anyone explain in detail??
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by rijul007 » Thu Dec 22, 2011 10:37 pm
No of even = No of odd

Total = 6
The are 3 of each type

Prob = 4C3*4C3/8C6 = 4*4/28 = 16/28 = 4/7

Option D
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by Anurag@Gurome » Fri Dec 23, 2011 12:10 am
karthikpandian19 wrote:In a particular lottery game, 6 numbers are chosen out of 4 even numbers and 4 odd numbers. What is the probability that the six 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

Can anyone explain in detail??
Total # of ways = 8C6
# of ways of choosing 3 even and 3 odd numbers = 4C3 * 4C3
Therefore, required probability = (4C3 * 4C3) / 8C6 = [spoiler]4/7[/spoiler]

The correct answer is D.
Anurag Mairal, Ph.D., MBA
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by ArunangsuSahu » Fri Dec 30, 2011 11:51 am
4C3*4C3=16=Event

Sample Space = 8C6=28

P=16/28=4/7
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