If n integers taken at random are multiplied together, then the probability that the last digit of the product is 1,3, 7 or 9 is

A
$2^{n} / 5^{n}$
B
$4^{n} - 2^{n} / 5^{n}$
C
$4^{n} / 5^{n}$
D
None of these

Solution:

In any number of last digits can be 0, 1, 2, 3, 4, 6, 6,7,8, 9.
Therefore, last digit of each number can be chosen in 10 ways.
Thus, exhaustive number of ways is 10ⁿ.
if the last digit be 1, 3, 7 or 9, then none of the numbers can be even or end in 0 or 5.
Thus, we have a choice of four digits, viz. 1,3, 7 to 9 with which each of n numbers should end.
So, favorable number of ways is 4ⁿ.
Hence, the required probability is $\frac{4^{n}}{10^{n}} = {(\frac{2}{5})}^{n}$

Solution:

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