If 5 names are put on a ballot in random order, what is the probability they are not in alpahbetical order?

Answer 1

If we specify that no two names are identical, the answer is

1 - [1 / 5!] = 99.17%

Answer 2

The easiest way to calculate this is to first find the probability that all the names are in the correct order. So, the chances for the first name is 1 in 5. The chances for the second name are 1 in 4 etc. So we have 1/5 * 1/4 * 1/3 * 1/2 * 1/1 = .00833

So the chances of them not being in alphabetical order would be 1 - .00833 = .99167

The probablility the names will not be in alphabetical order would then be 99.17%

Answer 3

Probabability that the first name is in order= 26/26=1 Probabability that the second name is in order= 1/26 Probabability that the third name is in order= 1/26 ...... answer: Five names are put on a ballot in random order. What is the probability they are in alphabetical order=(1/26)^5 Five names are put on a ballot in random order. What is the probability they are not in alphabetical order=1-(1/26)^5 1-(1/26)^5=0.99999991 therefore, the probability is 99.999991%

Answer 4

im 99% sure that the answer is 119/120 because there are 5 possibilities for the first name, 4 possibilities for the second name, 3 possibilities for the third name, 2 possibilities for the fourth name, and 1 possibility left for the last name. 5x4x3x2x1 or 5! (5 factorial) = 120 different orders, only one is alphabetical so 119 orders out of 120 are not alphabetical.

Answer 5

5! = 120 ways to arrange the 5 names, only 1 of which is alphabetical order, so the probability is 1/120.