Explanation

(c)
The last digit of the product will be 1, 2, 3, 4, 5, 6, 7, 8 or 9 if and only if each of the $n$ positive integers ends in any of these digits. Now the probability of an integer ending
$in\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8\ or\ 9\ is\frac{8}{10}.Therefore\ the\ probability\ that\ the\ last\ digit\ of\ the\$
product of $n$ integer in $1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8\ or\ 9\ is\ \left( \frac{4}{5} \right)^{n}.$The probability for an integer to
end in $1,\ 3,\ 7$ or $9\ is\frac{4}{10} = \frac{2}{5}$
Therefore the probability for the product of $n$ positive integers to end in 1, 3, 7 or 9 is $\left( \frac{2}{5} \right)^{n}$
Hence the required probability$= \left( \frac{4}{5} \right)^{n} - \left( \frac{2}{5} \right)^{n} = \frac{4^{n} - 2^{n}}{5^{n}}$