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Difficult Math Problem #111 - Probability

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Difficult Math Problem #111 - Probability

by 800guy » Mon Mar 19, 2007 9:44 am
There are 8 members; among them are Kelly and Ben. A committee of 4 is to be chosen out of the 8. What is the probability that Ben is chosen to be in the committee and Kelly is not?

from diff math doc, oa coming when people respond with explantions
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by Neo2000 » Mon Mar 19, 2007 5:00 pm
Number of ways of picking 8people = 8C4

If Ben is in the committe, then 3places have to be filled from 7people. However Kelly cannot be in the committee so now 3 places have to be filled from 6people

Probability = (1x6C3)/8C4

??
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by jayhawk2001 » Mon Mar 19, 2007 9:16 pm
Neo, your answer sounds appealing. However, I can't seem to explain
one anomaly :-)

Slightly expanding on the question -- we have to form 2 committees of
4 members each: 1 with Ben + 3 members and another with Kelly + 3 members (the leftover if you will).

If we use the logic you have provided, we will end up with a probability of
6C3 / 8C4 = 20 / 70 = 2/7

So, the other committee with Kelly in it (that follows the same logic
that Ben should not be in it) will hence have the same probability 2/7.

Sum of probabilities does not add up to 1. Please correct me if I have
missed something obvious.

I was initially thinking that it would be 7C4 / 8C4 (yielding 1/2).
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by jayhawk2001 » Mon Mar 19, 2007 9:31 pm
Ah, I now see why the probabilities don't add up to 1.

The 3 groups are (Ben + 3) + (Kelly + 3) + (Ben + Kelly + 2)

(1C1*6C3 / 8C4) + (1C1*6C3 / 8C4) + (2C1*6C2 / 8C4) = 1

My 7C4 / 8C4 is incorrect as 7C4 includes cases where Ben is not in
the committee.
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by Neo2000 » Mon Mar 19, 2007 9:32 pm
Since we are originally forming only 1 committee you've missed out the group where Ben and Kelly both ARE in the group

If both are in the group then it would become 6C2/8C4 = 3/7
Slightly expanding on the question -- we have to form 2 committees of
4 members each: 1 with Ben + 3 members and another with Kelly + 3 members (the leftover if you will).
In this case, the second case cannot be 2/7 since only 4 people are left! If 4 people are left AND you are trying to form a committe of 4, the probability of that is 1.

Again the key word is "and" When you see "and" in probability you know that it indicates multiplication. so you should have been multiplying NOT adding.

Hope this helps
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by Cybermusings » Tue Mar 27, 2007 10:04 am
Neo's explanation sounds appealing...Whats the OA?
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oa

by 800guy » Thu Mar 29, 2007 10:34 pm
oa:

let’s assume Ben has already been chosen. Then I have to choose 3 more people from the remaining, excluding Kelly, that is, three from six people, that’s 6c3.
so the total is (1c1.6c3)/8c4 which is 2/7
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Re: Difficult Math Problem #111 - Probability

by kevincanspain » Fri Mar 30, 2007 3:23 pm
800guy wrote:There are 8 members; among them are Kelly and Ben. A committee of 4 is to be chosen out of the 8. What is the probability that Ben is chosen to be in the committee and Kelly is not?

from diff math doc, oa coming when people respond with explantions
Let B be the event that Ben is chosen and K that Kelly is not chosen

We are to find P(B and K)= P(B)*P(K/B)=1/2*4/7=2/7

Alternatively, n(B and K)= 6C3=20
n(S)= 8C4= 8*7*6*5/24=70

P(B and K)=20/70=2/7
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