+0

# help me quick!

0
635
6
+4325

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?

Dec 13, 2017

#2
+105533
+2

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

.
Dec 13, 2017
edited by Melody  Dec 13, 2017
edited by Melody  Dec 13, 2017
edited by Melody  Dec 13, 2017
#5
+28179
+2

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.

Dec 13, 2017
#6
+105533
+1

Thanks Alan

Melody  Dec 14, 2017