There are only two women among $20$ persons taking part in a pleasure
trip. The $20$ persons are divided in two groups, each group consisting
of $10$ persons. Then the probability that the two women will be in the
same group is ?
Attempt:
$n(S) = \dfrac{20!}{2!10! 10!}$
$n(E) = \dfrac{19!}{10! 9!}$ (considering the two women as one unit)
^(I am dividing 19 into groups of 9 and 10)
$P(E) = \dfrac{n(E)}{n(S)} = 1$
which is obviously wrong.
What is my error?
Best Answer
What we have to do here is that we need to put the $2$ women in same group.
So firstly,
$n_s=$ # of ways in which 20 people can be distributed in group of 2 tens $= 20!/(10!10!2!)$
$n_e=$# of ways in which 18 people can be divide into group of 10 and 8 $= 18!/(10!8!)$
$P(E)=n_e/n_s=(18!/(10!8!))*((10!10!2!)/20!)=10*9*2/(20*19)=9/19$
If we consider the 2 women as one unit, then we in whichever group they lie we can only put 8 more people. If we consider it your way then in the case when the unit of women lie in group of 9 people(possible since we are assuming them to be one unit), then we will have 11 people in total in that group which is violation of problem statement. So instead we divide the group of 18 people into 10 and 8 and then put the women in group of 8.