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Twelve balls are distributed among three boxes. The probability that the first box contains three balls is
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a
11092310
b
91102310
c
12C3123×29
d
12C3312
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Correct option is A
Since each ball can be placed in any one of the 3 boxes, therefore there are 3 ways in which a ball can be placed in any one of the three boxes. Thus, there are 312 ways in which 12 balls can be placed in 3 boxes. The number of ways in which 3 balls out of 12 can be put in the box is 12C3 The remaining 9 balls can be placed in 2 boxes in 29 ways. So, required probability is 12C331229=11092310Similar Questions
The odd against an event are 5 to 2 and the odds in favour of another disjoint event are 3 to 5. Then the probability that atleast one of the events will happen is
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