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?