# 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

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 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. PamFound 2 solutions by edjones, CharlesG2:Answer by edjones(7569)   (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(828)   (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 %