A teacher will pick a group of 4 students from a group of 8 students that includes Bart and Lisa. If one of all 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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Easier to solve using combinatorics.
ReplyDelete8c4 = 8.7.6.5/24 = 70
ReplyDelete6C2 = 15
15/70 = 3/14
Why 6C2 and not 4C2?
ReplyDeleteFelix, here is my thinking.
ReplyDeleteSplit 8 students into two groups: (1) 6 students without bart n lisa; (2) bart n lisa
In how many ways can one select 4 students among 8 students in such a way that bart n lisa are included?
Select 2 from (lisa, bart); select two from the remaining 6.
Total = 2C2*6C2 = 1*15 = 15
Selecting 4 out of 8 = 8C4 = 70
The answer would be = 15/70 = 3/14.
But Ariel says it is D. So I am confused as well.
I think it should be B .... 3/14
ReplyDeleteAns.
Probability of 2 fixed people getting selected = (combination of rest of 2 to be selected)/(Combination of all 4 to be selected)
= 6C2 / 8C4 = 3/14
Total number of different ways by which 4 people can be selected from a group of 8 is 8C4 = 70
ReplyDeletenow we know L and B will be there and the task is to find the remaining two from 6
L B _ _
This can be found using 6 C 2= 15
Probability = 15/ 70 = 3/14
Thanks for sharing
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