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Question

The number of possible outcomes in a throw of $$n$$ ordinary dice in which at least one of the dice shows an odd number is


A
6n1
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B
3n1
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C
6n3n
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D
none of these
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Solution

The correct option is C $$6^{n} - 3^{n}$$
Total number of possible outcomes in a throw of $$n$$ ordinary dies $$=6\times 6\times 6\times 6\times .................\quad n-times={ 6 }^{ n } $$
The even numbers in the throw of an ordinary dice are $${2,4,6}$$
The total number of possible outcomes when no dice show odd number $$=3\times 3\times 3\times 3\times .............n-times={ 3 }^{ n }$$
Hence the number of possible outcomes in which at least one dice shows an odd number$$=$${Total possible outcomes-possible outcomes when no dice shows an odd number}
$$={ 6 }^{ n }-{ 3 }^{ n }$$
Hence, option 'C' is correct.

Maths

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