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Only Jim and not Jill

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wrong calculation

by 007.r.mason » Sat Feb 28, 2009 7:09 pm
I think 8C3 is 56 not 48...
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by sanju09 » Mon Mar 02, 2009 4:20 am
DanaJ wrote:The wording of your problem makes me wonder what you actually mean. On a first glace (and I'm probably not going to give it a second glance, since I HATE probabilities), you're looking for (the number of 3 person groups containing Jack but not Jill)/(total number of groups of 3 people)
The total number of groups will be 8C3 = 56.
Groups that contain Jack have "two open positions" for the other 6 people. There are 6C2 possible groups of two people from the pool of 6 left or 15.
This makes the solution to this problem (IMHO, of course, since I DESPISE probabilities) 15/56.
:) So, Jack is also in the pool, great info DanaJ!
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by keeyu2 » Thu Mar 12, 2009 6:34 am
I think the answer is 5/56.

The proability of Jim is 1/8
The proability of not Jill is 6/7
The proability of not Jill again is 5/6

These are dependent P(Jim & Not Jill & Not Jill).

1/8*6/7*5/6 is 5/56.


What is the OG answer?
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by Reader » Sun Mar 15, 2009 10:34 am
I think it's 15/56,

the chance of jim being select = 3/8, the chance both jim and jill being selected = 3/8 x 2/7 =6/56. only jim not jill = 3/8 - 6/56 = 15/56
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