Submitted by Nasiful on Tue, 08/29/2017 - 12:27
Forums:
At the end, they have the STEP question on probability but I was unable to actually find a solution which uses the factorial method instead of the 'n choose r' method, if anyone has a solution that works, please post it;
Question is as follows, I seat n boys and 3 girls in a line at random, so that each order of the n + 3 children is as
likely to occur as any other. Let K be the maximum number of consecutive girls in the line
so, for example, K = 1 if there is at least one boy between each pair of girls.
i) Find P(K=3)
(ii) Show that
P(K = 1) = n(n − 1)/((n + 2)(n + 3))
(iii) Find E(K)
Thanks in advance
Please can you tell us what
Please can you tell us what you've attempted so far so that we can give you the most useful hint(s). I'm unsure what you mean by 'factorial method' vs 'n choose r' method but I'd like to point out that any valid method is acceptable.
Method that I tried
So they wanted the maximum ways that the girls could all be together, so of course all the different arrangments are (n+3)! and it should be divided by 3! because there are 3 girls right? but where do I go from here?
Ok, so in this question you
Ok, so in this question you have to use the fact that each permutation of the boys and girls is equally likely so to calculate probabilities you count the total number of permutations that correspond to the event you want to calculate the probability for and divide by the total number of permutations. This is called elementary probability.
You have correctly identified $(n+3)!$ as the total number of permutations. For the first part, the number of permutations corresponding to $K=3$ can be calculated by considering the group of 3 girls as 1 child as far as the permutation is concerned so this means there are $n+1$ children to permute. You then just need to take into account the fact that the girls can permute within their group of 3 so multiply what you have by $3!$. Can you see what the answer to the first part is now?
Yes actually, thank you so
Yes actually, thank you so much