Help maths problem. I am stuck.
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Help maths problem. I am stuck.

[From: ] [author: ] [Date: 12-08-27] [Hit: ]
I have obviously tried this, but there is not point in me showing you my wrong working out as it is quite long. I hope someone can help me here...answer is 4/27-there are 4 odd #s & 8 even #s,......
In a 3-ball lottery, three numbered balls are released from a chamber. the number of even-numbered balls in the chamber is twice the number of odd numbered balls. what is the probability that at least 2 of the 3 released balls will be odd? assume that there are 12 in the chamber and balls are released without replacement.

I have obviously tried this, but there is not point in me showing you my wrong working out as it is quite long. I hope someone can help me here...

answer is 4/27

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there are 4 odd #s & 8 even #s, total 12
we want {2 odd, 1 even} or {3 odd}
use combinations

Pr = (4c2*8c1 + 4c3)/12c3 = 52/220 = 13/55 <-------

the given ans is wrong. let's doublecheck with direct probabilities
3* 4/12 *3/11 *8/10 + 4/12 *3/11 *2/10 = 13/55 <-------

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im bad at math
1
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