A group of individuals containing 'b' boys and 'g' girls is lined up in random order. Assuming all permutations are equally likely, what is the probability that the person in the i-th position is a girl, 1<=i<=g? (i is more than or equal to 1 and less than or equal to g)
find probability that i-th person in line is female
- Thread starter serious
- Start date
- Joined
- Feb 4, 2004
- Messages
- 16,583
Is perhaps the exercise providing an additional condition for consideration...?Denis said:
We have g + b people in line, so i should clearly be no more than g + b for the exercise to make sense. So this may be an unstated condition. The exercise may specify i < g to clarify that it is asking for the probability that, considering the first g places in line (from the first person to the g-th), what is the probability that the i-th person is female?
I'm just guessing, though....
Eliz.
Hey thanks for replying. And yup its really 1<=i<=g, meaning that the i-th position is confined to the first g places. And if its helpful, the assumption is that all permutations are equally likely. All arrangement of boys and girls are equally likely.
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
It may come as a surprise to many but it makes no difference as to the value if i as long as \(\displaystyle 1 \leqslant i \leqslant g + b\) the answer is the same.serious said:
There are \(\displaystyle {\left( {b + g} \right)!}\) ways to arrange this group.
There are \(\displaystyle {g \cdot \left( {b + g -1} \right)!}\) ways to arrange this group with a girl in the ith position.
So what is the probability?