Any clever tell me the probability in the following situation?

Question

I have a box with 100 balls each having a unique number from 1 to 100. what chance do I have to draw 100 balls coming out in the correct order 1, 2, 3, 4,...,99,100. And please state the equation used to calculate. thanks in advance.

Answer 1

1/100 * 1/99 * 1/98 *........*1/2 *1 do the math yourself.
each ball has equal probablility of being taken out. take the first one out there are 99 left again with equal probability, and so on

Answer 2

I believe this is called permutation. The chance of doing so is 100! (the exclamation point means that the equation will be 100 x 99 x 98 x ... x 3 x 2 x 1). My Excel spreadsheet won't fit all the digits, but if that's the correct equation, the answer is 9.3326E+157. The formula to use for Excel would be =permut(100,100).

Answer 3

The probability of drawing the balls in the correct order is exactly 1 in 93326 21544 39441 52681 69923 88562 66700 49071 59682 64381 62146 85929 63895 21759 99932 29915 60894 14639 76156 51828 62536 97920 82722 37582 51185 21091 68640 00000 00000 00000 00000 000

Formula: there are n! permutations of n balls, of which exactly one is correct. Thus the probability is 1/100!. That rather largish number there is 100! (if you're not familiar with the notation, n! = n*(n-1)*(n-2)*(n-3)... 3*2*1)

Answer 4

There will be aprox. 1 posibility in 2.200 X (10at150) (I don´t know how to write small numbers in this keyboard) and the equation is that you have 1 chance in 100 to take ball number 1, then 1 in 99 to take number 2, and so on...

Answer 5

You havn't given enough information. You havn't specified whether you replace each ball after it has been chosen.
Assuming you don't replace it, I think that to calculate it u do:
(1/100)x(1/99)x(1/98).....x(1/2)x1=your answer

Answer 6

ONLY 1