# SOLUTION: A bag contains 15 billiard balls, numbers consecutively from 1 to 15. Six balls are selected at random. What is the probability that the sum of the numbers on the billiard balls wi

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: A bag contains 15 billiard balls, numbers consecutively from 1 to 15. Six balls are selected at random. What is the probability that the sum of the numbers on the billiard balls wi      Log On

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 Question 143040: A bag contains 15 billiard balls, numbers consecutively from 1 to 15. Six balls are selected at random. What is the probability that the sum of the numbers on the billiard balls will be odd?Found 2 solutions by Edwin McCravy, edjones:Answer by Edwin McCravy(8999)   (Show Source): You can put this solution on YOUR website!``` The 7 even numbered balls are 2,4,6,8,10,12,14 The 8 odd numbered balls are 1,3,5,7,9,11,13,15. There will be an odd sum is there is an odd number of odd numbered balls, and an even sum if there is an even number of odd balls. So the cases of an odd number of odd numbered balls: N(1 odd numbered ball & 5 even numbered balls) = C(8,1)*C(7,5) = 8*21 = 168 N(3 odd numbered ball & 3 even numbered balls) = C(8,3)*C(7,3) = 56*35 = 1960 N(5 odd numbered ball & 1 even numbered balls) = C(8,5)*C(7,1) = 56*7 = 392 Total = 2520 The total number of choices of any 6 balls out of the 15 is C(15,6) = 5005 So the desired probability = 2520/5005 = 72/143 Edwin``` Answer by edjones(7569)   (Show Source): You can put this solution on YOUR website!8 are odd, 7 are even. All the possible combinations are: 8C6=28 [all odd] 8C5 * 7C1 =56*7=392 [5 odd, 1 even] 8C4 * 7C2 =70*21=1470 [4 odd, 2 even] 8C3 * 7C3 =56*35=1960 [3 odd, 3 even] 8C2 * 7C4 =28*35=980 [2 odd, 4 even] 8C1 * 7C5 =8*21=168 [1 odd, 5 even] 7C6=7 [all even] . Adding all these numbers we get 5005 possible combinations. Only those with an odd number of odd numbers give an odd result. They are: 392+1960+168=2520 2520/5005=72/143 or .503497 Probability that the sum of the numbers will be odd. . Ed