From a group of 8 people that includes George and Marge we select a group of 3. What is the probability that the group contains George and doesn't contain Marge?
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From a group of 8 people that includes George
Source: Beat The GMAT — Problem Solving |
I think it's 15/56.
Total number of groups of 3 persons to be selected out of 8: 8!/(3!*5!)= 56
Total number of groups containing George = number of posibilities to selected 2 people out of 7 (1 place is taken by George): 7!/(2!*5!)= 21.
Total number of groups containing George AND Marge: 6!/(1!*5!)= 6.
Total number of groups containing George but not containg Marge: 21 - 6 = 15.
Probability of selecting a group that contains George but does not contain Marge is then: 15/56
Total number of groups of 3 persons to be selected out of 8: 8!/(3!*5!)= 56
Total number of groups containing George = number of posibilities to selected 2 people out of 7 (1 place is taken by George): 7!/(2!*5!)= 21.
Total number of groups containing George AND Marge: 6!/(1!*5!)= 6.
Total number of groups containing George but not containg Marge: 21 - 6 = 15.
Probability of selecting a group that contains George but does not contain Marge is then: 15/56
15/56
no ways of making a group 3 person out of 8 is 8c3=8!/(3!*5!)=56
now in a group of 3 george is fix , for other two member we have to make selection of 2 from 6 as marge is not to be selected => 6c3=6!/(3!*2!)=15
probabilty=15/56
no ways of making a group 3 person out of 8 is 8c3=8!/(3!*5!)=56
now in a group of 3 george is fix , for other two member we have to make selection of 2 from 6 as marge is not to be selected => 6c3=6!/(3!*2!)=15
probabilty=15/56
It does not matter how many times you get knocked down , but how many times you get up
Can you explain the second step:
"now in a group of 3 george is fix , for other two member we have to make selection of 2 from 6 as marge is not to be selected => 6c2=6!/(2!*4!)=15"????
"now in a group of 3 george is fix , for other two member we have to make selection of 2 from 6 as marge is not to be selected => 6c2=6!/(2!*4!)=15"????
I calculated:
Choose George not Marge: 2C1=2
Choose 2 of the remaining 6 persons: 6C2=15
Choose a group of 3 out of 8 persons: 8C3=56
2*15/56=30/56=15/28
Choose George not Marge: 2C1=2
Choose 2 of the remaining 6 persons: 6C2=15
Choose a group of 3 out of 8 persons: 8C3=56
2*15/56=30/56=15/28
15/56
8C3 is total # of combinations = 56
If you choose G then 1 place is reserved. You have to choose 2 more ppl from 6 remaining (no Marge)
= 6C2 = 15
So probability = 15/56
8C3 is total # of combinations = 56
If you choose G then 1 place is reserved. You have to choose 2 more ppl from 6 remaining (no Marge)
= 6C2 = 15
So probability = 15/56
Saw a similar question on Exam 5, except it was George and Nina (as opposed to George and Marge). I know this is an old thread. Just posting to add that the solution can also be found without resorting to combinatorics, by applying principles of probability. The probability that George is selected is 3/8 (that is there are 8 candidates to fill three slots at random). And the probability that Marge (or Nina) is not selected, given that George is selected, is 5/7. That is, there are 7 candidates to fill the remaining two slots (meaning 5 will not be selected). The overall probability of both events occurring is thus 3/8 x 5/7 = 15/56.
800 or bust!
You can also imagine selecting people one at a time. There are three options for getting what we want:
George, not Marge, not Marge --> (1/8)(6/7)(5/6)
not Marge, George, not Marge --> will end up same as above
not Marge, not Marge, George --> will end up same as above
so 3 ways each with a (1/8)(6/7)(5/6) probability. So 3*(1/8)(6/7)(5/6) = 15/56
George, not Marge, not Marge --> (1/8)(6/7)(5/6)
not Marge, George, not Marge --> will end up same as above
not Marge, not Marge, George --> will end up same as above
so 3 ways each with a (1/8)(6/7)(5/6) probability. So 3*(1/8)(6/7)(5/6) = 15/56
