If 12 identical balls are to be placed in 3 different boxes, then the probability that one of the boxes contains excatly 3 balls, is :

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Solution

Total outcomes =
${(12, 0, 0), (11, 1, 0), (10, 1, 1), (10, 2, 0)$
$(9, 3, 0), (9, 2, 1), (8, 4, 0), (8, 3, 1)$
$(8, 2, 2), (7, 5, 0), (7, 4, 1)$
$(7, 3, 2), (6, 6, 0), (6, 5, 1),$
$(6, 4, 2) (6, 3, 3) (5, 5, 2) (5, 4, 3)$
$(4, 4, 4)} = 19$
Favorable cost = 4
${(3, 9, 0), (3, 8, 1), (3, 7, 2), (3, 5, 4)}$
Probability = $4 / 19$

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If 12 identical balls are to be placed in 3 different boxes, then the probability that one of the boxes contains excatly 3 balls, is :

Total outcomes = {(12,0,0),(11,1,0),(10,1,1),(10,2,0)(9,3,0),(9,2,1),(8,4,0),(8,3,1)(8,2,2),(7,5,0),(7,4,1)(7,3,2),(6,6,0),(6,5,1),(6,4,2)(6,3,3)(5,5,2)(5,4,3)(4,4,4)}=19Favorable cost = 4 {(3,9,0),(3,8,1),(3,7,2),(3,5,4)}Probability = 4/19