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
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