Thread: Probability- pick 10 numbers from 1-100

Hi all,

You choose a number at random from the numbers 1-100 ten times. What is the probability of choosing 10 distinct numbers? What is the probability that the first number chosen is larger than each of the other nine numbers chosen?


So, for the first question, I compute the following- (100*99*98*97*96*95*94*93*92*91)/(100^10) = .628

For the second question, I can either first draw a one, OR a two, OR a three. If I draw a 1, nothing is lower so the probability is 0. If I draw a 100, 99 numbers are lower. So, I perform the following calculation-

SUM (1/100) * ((n/100)^9) for n = 1 to 99.

This is equivalent to saying "probability I first draw number n AND I subsequently draw 9 numbers lower than it"

I get .095. Is this correct? Is there a better way?

Originally Posted by sfspitfire23

Hi all,

You choose a number at random from the numbers 1-100 ten times. What is the probability of choosing 10 distinct numbers? What is the probability that the first number chosen is larger than each of the other nine numbers chosen?
So, for the first question, I compute the following- (100*99*98*97*96*95*94*93*92*91)/(100^10) = .628
For the second question, I can either first draw a one, OR a two, OR a three. If I draw a 1, nothing is lower so the probability is 0. If I draw a 100, 99 numbers are lower. So, I perform the following calculation-
SUM (1/100) * ((n/100)^9) for n = 1 to 99.
This is equivalent to saying "probability I first draw number n AND I subsequently draw 9 numbers lower than it"
I get .095. Is this correct? Is there a better way?

The reason I deleted my reply is that it is not clear to me from the wording of the question if the numbers are distinct as in part #1 or not.

I do agree with you on that part.

Are they distinct or not?

The numbers do not have to be distinct in the second part of the question