Question

The probability of an event under the uniform distribution - random permutations. A class with n kids lines up for recess. The order in which the kids line up is random with each ordering being equally likely. There are two kids in the class named Celia and Felicity. Give an expression for each of the probabilities below as a function of n. Simplify your final expression as much as possible so that your answer does not include any expressions of the form (ab)(ab).

(a) What is the probability that Celia is first in line?
(b) What is the probability that Celia is first in line and Felicity is last in line?
(c) What is the probability that Celia and Felicity are next to each other in the line?

Answer 1

a)

the probability that Celia is first in line is 1/n.

b)

the probability that Celia is first in line and Felicity is last in line is 1/n(n-1)

c)

the probability that Celia and Felicity are next to each other in the line is 2/n

Explanation:

Given the data in question;

a) Celia is first in line?

the total ways are; n! to place n kids in n places in orderly fashion.

now, if Celia is the first kid, we have to order remaining n-1 kids

(n-1)! / n! = 1/n

therefore, the required probability is 1/n.

b) Celia is first in line and Felicity is last in line?

the total ways are; n! to place n kids in n places in orderly fashion,

now, if Celia is the first and Felicity is the last kid, we order the remaining n - 2 kids, which can be done in (n-2)!

(n-2)! / n! = 1 / n(n-1)

therefore, the required probability is 1 / n(n-1)

c) Celia and Felicity are next to each other in the line

the total ways are; n! to place n kids in n places in orderly fashion,

now, if Celia and Felicity as single object, so that they can be put together, we arrange n-1 objects which can be done in (n-1)! ways.

But also these 2 can change the place so in total there are 2(n-1)! ways.

2(n-1)! / n! = 2/n

Therefore, the required probability is 2/n