Probability of Boys and Girls in a Randomly Lined Up Group: Permutation Question

  • Thread starter Thread starter chocok
  • Start date Start date
  • Tags Tags
    Permutation
AI Thread Summary
The discussion revolves around calculating probabilities in a lineup of 5 boys and 10 girls. For part a), the probability that the person in the 4th position is a boy is determined by comparing the number of favorable outcomes to the total permutations, leading to a probability of 1/3. For part b), the probability that a specific boy is in the 3rd position is based on the total number of individuals, resulting in a straightforward calculation. Participants emphasize understanding the basic principles of probability rather than focusing on complex permutations. The key takeaway is that the position does not affect the overall probability when selecting randomly from the group.
chocok
Messages
21
Reaction score
0
A group of 5 boys and 10 girls is lined up in random order-that is, each of the 15! permutaions is assumed to be equally likely.

a) What is the probability that the person in the 4th position is a boy?
b) What is the probability that a particular boy is in the 3rd position?

i tried to list out 1 possible combination but it gets pretty ugly since there are a total of 15 choices.. can someone please help?
 
Physics news on Phys.org
chocok said:
A group of 5 boys and 10 girls is lined up in random order-that is, each of the 15! permutaions is assumed to be equally likely.

a) What is the probability that the person in the 4th position is a boy?
b) What is the probability that a particular boy is in the 3rd position?

i tried to list out 1 possible combination but it gets pretty ugly since there are a total of 15 choices.. can someone please help?

I'm not sure about this, so I'll take a guess at what I think.

For a), wouldn't you just find the number of possible ways to organize the line such that the 4th position is a boy? Once you got that, you can just compare it with the total permutation, which should give you the probability of that happening.

For b), it's something extremely similiar, but you have to remember that you cannot change position 3.

Is your problem with the idea of finding the number of ways to organize the line with some conditions?

If that's the case, try googling some sample exercises about permutations. After a couple, you'll understand enough to apply it to this situation. Practice is what makes perfect.
 
Last edited:
Thanks for replying.

for part a), If I just look at the first 3 picks, i can either have 0,1,2,3 boys and the 4th postion have to be a boy. I calculated the probabilities of each of them individually and they added up to be 1/3. For which i multiplied it by 11! (5-15th selection) and divided by 15! (total permutation), but i ended up with 1.0175x10^-5 which seems way to small to be a correct answer..
what did i do wrong in my reasoning??
 
The precise position isn't relevant. What may be confusing you is that "permutations" are not at all relevant. Suppose, out of 5 boys and 10 girls, you choose one person at random (each equally likely to be picked). What is the probability that the person is a boy? (That should be easy.)

"What is the probability that a particular boy is in the third position."
Given that the person in the third position is a boy?? Or given that the person in the fourth position was a boy? If neither of those is intended: there are 15 persons who are all equally likely to be chosen. What is the probability that a specific one of them will be chosen?

If it is intended that the person in the third position is a boy, then there are 5 persons all equally likely to be chosen.
 
Halls of Ivy: The precise position isn't relevant.

I remember an old Mutt and Jeff cartoon. As they are traveling on a train, Jeff asks Mutt, "How many sheep do you think are in that herd we just passed?"

Mutt replies, "876."

"That's amazing! You mean you counted every one as the train sped past?"

"No, not really. I just counted the legs and divided by four."
 
Last edited:
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top