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?
Teacher's Pets (probability)
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
-
xcusemeplz2009
- Master | Next Rank: 500 Posts
- Posts: 399
- Joined: Wed Apr 15, 2009 3:48 am
- Location: india
- Thanked: 39 times
-
Andrei
- Junior | Next Rank: 30 Posts
- Posts: 28
- Joined: Wed Feb 14, 2007 12:25 am
- Thanked: 1 times
- GMAT Score:690
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.
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.
