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
B
3n1
C
6n3n
D
none of these

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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