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RedeemOnly Jim and not Jill
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007.r.mason
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sanju09
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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!The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
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?
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?