The probability that when 12 distinct balls are distributed among three boxes, the first will contain exactly three balls is
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a
29312
b
12C3⋅2996
c
12C3⋅212312
d
12C3⋅39312
answer is B.
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Detailed Solution
3 balls for 1st Box can be selected in 12C3 ways, remaining 9 balls can be placed in remaining 2 boxes in 29 ways.Number of favourable cases =12C3×29Total number of cases = 312Required Probability = 12C3×29312= 12C3×2996