+4622 If 8 numbers are chosen at random from the first 15 positive integers, what is the probability that an additional number chosen at random from the remaining 7 numbers is less than any of the 8 previously chosen numbers?
+118687 If 8 numbers are chosen at random from the first 15 positive integers, what is the probability that an additional number chosen at random from the remaining 7 numbers is less than any of the 8 previously chosen numbers?
I just did this in detail but my internet went down and the whole lot got lost.
I'll try to recap.
Prob = prob when biggest number is 9 + prob when biggest number is 10 ...... prob when biggest number is 15
=(8C7 / 15C8) * (1/7)
+(9C7 / 15C8) * (2/7)
+(10C7 / 15C8) * (3/7)
+(11C7 / 15C8) * (4/7)
+(12C7 / 15C8) * (5/7)
+(13C7 / 15C8) * (6/7)
+(14C7 / 15C8) * (7/7)
=(8C7+9C7*2+10C7*3+11C7*4+12C7*5+13C7*6+14C7*7) / (15C8*7)
nCr(8,7)+nCr(9,7)*2+nCr(10,7)*3+nCr(11,7)*4+nCr(12,7)*5+nCr(13,7)*6+nCr(14,7)*7 = 40040
nCr(15,8)*7 = 45045
40040/45045 = 0.8888888888888889
Probability that the single number chosen is smaller than at least one of the other 8 already chosen
\(=\frac{8}{9}\)
Maybe LOL 
+33661 I think the question is ambiguous. Does "any" mean "all", or does it mean "at least one"?
I've done some MonteCarlo simulations using 10^7 trials.
(1) Assuming "any" means "at least one", I get a probability of 0.889 (to 3 significant figures).
Melody found 8/9 ≈ 0.889
(2) Assuming "any" means "all", I get a probability of 0.111 (to 3 significant figures).
1/9 ≈ 0.111.
The logic of my MonteCarlo simulation is as follows:
1. Randomly permute the integers 1 to 15.
2. Test the integer in the ninth position against those in the first eight positions.
3. If the ninth integer is less than one or more of the first eight, score 1 for the "at least" sum; if it is less than all of the first eight, score 1 for the "all" sum.
4. Repeat steps 1 to 3 10^7 times.
5. Divide the sums by 10^7 to get the probabilities.
