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Help needed on probability question

Suppose that Mindy and John are part of a group of 40 students selected by the administration of the student council. Suppose/, furthermore, that this group was subdivided into two groups in such a way that 10 students were placed in subgroup#1 and 30 students in subgroup#2. What is the probability that Mindy and John were placed in the same subgroup?



So I know that the answer is the sum of the probabilities :
both placed in group 1 + both placed in group 2

For the probability that both of them are placed in Group 1, I know that
the total number of combinations is (4010)\displaystyle \binom{40}{10} but I don't know what the number of combinations with Mindy and John in it is..

can someone explain what it is please
(edited 10 years ago)
Reply 1
Avatar for Old_Simon
Old_Simon
12
Original post by crescendo7
So I know that the answer is the sum of the probabilities :
both placed in group 1 + both placed in group 2

For the probability that both of them are placed in Group 1, I know that
the total number of combinations is (4010)\displaystyle \binom{40}{10} but I don't know what the number of combinations with Mindy and John in it is..

can someone explain what it is please


The probability that one of them falls in the smaller group is 1/4.
So the probability that 2 of them are in it is (1/4)^2 ?

Can you go on from there ?
Reply 2
Avatar for crescendo7
crescendo7
OP
Original post by Old_Simon
The probability that one of them falls in the smaller group is 1/4.
So the probability that 2 of them are in it is (1/4)^2 ?

Can you go on from there ?



how would you solve it using permutations and combinations
The books says the number of combinations of Mindy and John being in Group1 is (388)\displaystyle \binom{38}{8} but I don't understand why
Original post by Old_Simon
The probability that one of them falls in the smaller group is 1/4.
So the probability that 2 of them are in it is (1/4)^2 ?

Can you go on from there ?


That's not correct. Once one person is in the smaller group, the probability that the second one is also there is no longer 1/4, as there are only 9 slots out of the 39 remaining.

Original post by crescendo7
how would you solve it using permutations and combinations
The books says the number of combinations of Mindy and John being in Group1 is (388)\displaystyle \binom{38}{8} but I don't understand why


If Mindy and John are in the smaller group, then there are 8 slots left in the that group, and there are 38 people left to choose from. Hence the result.
Reply 4
Avatar for Old_Simon
Old_Simon
12
Original post by crescendo7
how would you solve it using permutations and combinations
The books says the number of combinations of Mindy and John being in Group1 is (388)\displaystyle \binom{38}{8} but I don't understand why

Well as a starter u need the probability of both being in the small group. Then go from there. You need the probability of them both being in the larger group. Then you do what ?
Reply 5
Avatar for davros
davros
16
Original post by crescendo7
how would you solve it using permutations and combinations
The books says the number of combinations of Mindy and John being in Group1 is (388)\displaystyle \binom{38}{8} but I don't understand why


If Mindy and John are already given to be in Group 1 then you have to choose another 8 people for that group from the remaining 38 people.
Reply 6
Avatar for crescendo7
crescendo7
OP
thanks

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