Question

Exercise 12.1.2: The probability of an event under the uniform distribution - random permutations. About 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). (a) What is the probability that Celia is first in line?

Answer 1

Answer:

a) $P_a=\frac{1}{n}$

b) $P_b=\frac{1}{n(1-n)}$

c) $P_c=\frac{2}{n}$

Step-by-step explanation:

The question is incomplete:

(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?

a) The probability that Celia is first in line, if there are n kids and all of them have the same chance, is 1/n.

$P_a=\frac{1}{n}$

b) The probability that Celia is first in line and Felicity is last in line is

$P_b=\frac{1}{n} \frac{1}{n-1} =\frac{1}{n(1-n)}$.

It is deducted like that:

If Celia is placed in the first line (what has a probablity of 1/n), there are left (n-1) kids. Then, the probability of placing Felicity in the last place is 1/(n-1).

Both probabilities multiplied give 1/(n*(n-1)).

c) The probability that Celia and Felicity are next to each other in the line is

$P_c=\frac{2(n-1)!}{n!}=\frac{2}{n}$

There are n! combinations for kids lines, where (n-1)! permutations have Celia before Felicity and other (n-1)! permutations have Felicity before Celia.

Then, we have 2(n-1)! permutations that have Celia and Felicity next to each other, over n! permutations possible.