document.write( "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 \n" ); document.write( "
Algebra.Com's Answer #266655 by CharlesG2(834)\"\" \"About 
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\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "999 numbers
\n" ); document.write( "700 to 999 is 300 numbers
\n" ); document.write( "P(r) = C(n,r) * p^r * q^(n-r)
\n" ); document.write( "q = 1 - p
\n" ); document.write( "C(n,r) = n!/(r! * (n-r)!)
\n" ); document.write( "r successes in n tries
\n" ); document.write( "C(10,7) = 10!/(7! * (10 - 7)!)
\n" ); document.write( "C(10,7) = 10!/(7! * 3!)
\n" ); document.write( "C(10,7) = (8 * 9 * 10)/(1 * 2 * 3)
\n" ); document.write( "C(10,7) = 720/6 = 120
\n" ); document.write( "p^r = (300/999)^7
\n" ); document.write( "q^(n - r) = (699/999)^3
\n" ); document.write( "P(7) = 120 * (300/999)^7 * (699/999)^3
\n" ); document.write( "0.00905329504280360257128551493191386
\n" ); document.write( "P(7) = 0.009053 rounded to 6 places
\n" ); document.write( "P(7) = 0.9053 %\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );