Question

If $n$ positive integers are taken at random and multiplied together, the probability that the last digit of the product is $2,4,6$ or 8 , is

• A. $\frac{4^n+2^n}{5^n}$
• B. $\frac{5^n-2^n}{5^n}$
• C. $\frac{4^n-2^n}{5^n}$
• D. $\frac{4^n-3^n}{5^n}$

Answer 1

Answer: (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 $n$ integers in $1,2,3,4,6,7,8$ 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 ${\left(\frac{2}{5}\right)}^n$.
$={\left(\frac{4}{5}\right)}^n-{\left(\frac{2}{5}\right)}^n=\frac{4^n-2^n}{5^n}$