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by mlaboda » Mon Sep 07, 2009 3:05 pm
A teacher will pick a group of 4 students from a group of 8 students that includes Bart and Lisa. If one of all of the possible four-student groups is picked at random, what is the probability of picking a group that includes both Bart and Lisa?
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IMO

by xcusemeplz2009 » Mon Sep 07, 2009 10:05 pm
is it 3/14
It does not matter how many times you get knocked down , but how many times you get up
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by Nermal » Tue Sep 08, 2009 12:08 am
I also got 3/14.

I calculated:
(2C2*6C2)/(8C4)
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IMO

by enniguy » Tue Sep 08, 2009 12:36 am
3/10
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by Andrei » Tue Sep 08, 2009 1:33 am
I also think it's 3/14.

Total number of possible groups of 4 persons selected out of 8 is: 8!/(4!*4!) = 70.

If both both Bart and Lisa must be in the group, than there are left only 2 places for the other 6 students, so the total number of possible groups including both Bart and Lista is equal to the number of combinations of 2 selecting persons out of 6: 6!/(4!*2!) = 15.

So the probability of selecting a group that includes Bart and Lista out of total groups of 4 students is: 15/70 = 3/14.
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