ProbabilityHard
Question
If 12 identical balls are to be placed in 3identical boxes, then the probability that oneof the boxes contains exactly 3 balls is :
Options
A.
B.
C.
D.
Solution
The actual solution corresponding to the language given in question will be
Total number of ways = n(s)
= Number of elements in sample space = 312
If E corresponds to the event that one of the boxes contains exactly 3 balls.
Then
n(E) = 12C3 × 3C1 × [29 - 2 × 9C3] + 12C6 × 3C1 × 6C3
∴ Required probability
Hence none of the answer matches from the given options so it seems that the question is ill-framed. If we assume the event as one of the particular box contains exactly 3 balls then
n(E) = 12C3.29
∴ Probability =
Total number of ways = n(s)
= Number of elements in sample space = 312
If E corresponds to the event that one of the boxes contains exactly 3 balls.
Then
n(E) = 12C3 × 3C1 × [29 - 2 × 9C3] + 12C6 × 3C1 × 6C3
∴ Required probability
Hence none of the answer matches from the given options so it seems that the question is ill-framed. If we assume the event as one of the particular box contains exactly 3 balls then
n(E) = 12C3.29
∴ Probability =
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