If a permutation is chosen at random, what is the probability that it begins with at least 2 A's?
The probability is
Incomplete
To find the probability that a permutation begins with at least 2 A's, we need to consider the total number of permutations and the number of permutations that satisfy the given condition.
Let's start by determining the total number of permutations. If we assume that our permutation consists of n distinct elements, the total number of possible permutations is given by n!.
Next, let's determine the number of permutations that begin with at least 2 A's. To count this, we will consider two cases:
Case 1: Exactly 2 A's at the beginning:
In this case, we fix the first two positions as A's and permute the remaining (n-2) elements. The number of permutations for this case is (n-2)!.
Case 2: More than 2 A's at the beginning:
In this case, we have fixed the first (k) positions as A's, where k > 2. We can choose k positions out of the n available positions, and permute the remaining (n-k) elements. The number of permutations for this case is given by choosing k positions out of n, which can be calculated using combinations notation: C(n, k). After choosing the positions, we permute the remaining (n-k) elements, which is represented as (n-k)!.
To find the total number of permutations with at least 2 A's at the beginning, we sum up the permutations from both cases:
Total permutations with at least 2 A's at the beginning =
(Number of permutations with exactly 2 A's at the beginning) +
(Number of permutations with more than 2 A's at the beginning)
Total permutations with at least 2 A's at the beginning =
[(n-2)!] + [C(n, 3) * (n-3)! + C(n, 4) * (n-4)! + ... + C(n, n) * (n-n)!]
Finally, we can calculate the probability by dividing the total number of permutations with at least 2 A's at the beginning by the total number of possible permutations:
Probability = (Total permutations with at least 2 A's at the beginning) / (Total number of permutations)