SOLUTION: If you have numbers from 1 to 999 in a bag and you pull out 10 numbers, what are the odds that 7 of those numbers will be from 700 to 999? Is there a formula for figuring this out

Question 374877: If you have numbers from 1 to 999 in a bag and you pull out 10 numbers, what are the odds that 7 of those numbers will be from 700 to 999? Is there a formula for figuring this out and if there is can you send it to me? Thanks for your help. Pam
Found 2 solutions by edjones, CharlesG2:
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
nCr=Combination of n things taken r at a time=n!/((n-r)!r!)
(300C7 * 699C3)/999C10 = .0088 = 88/10,000 Probability that 7 of those numbers will be from 700 to 999.
10000-88=9912
Odds against 7 of those numbers will be from 700 to 999 = 9912 to 88 or 113 to one. The odds in favor are the reverse, one to 113.
.
Ed
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
If you have numbers from 1 to 999 in a bag and you pull out 10 numbers, what are the odds that 7 of those numbers will be from 700 to 999? Is there a formula for figuring this out and if there is can you send it to me? Thanks for your help. Pam

999 numbers
700 to 999 is 300 numbers
P(r) = C(n,r) * p^r * q^(n-r)
q = 1 - p
C(n,r) = n!/(r! * (n-r)!)
r successes in n tries
C(10,7) = 10!/(7! * (10 - 7)!)
C(10,7) = 10!/(7! * 3!)
C(10,7) = (8 * 9 * 10)/(1 * 2 * 3)
C(10,7) = 720/6 = 120
p^r = (300/999)^7
q^(n - r) = (699/999)^3
P(7) = 120 * (300/999)^7 * (699/999)^3
0.00905329504280360257128551493191386
P(7) = 0.009053 rounded to 6 places
P(7) = 0.9053 %

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