Question

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

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

Answer 1

The correct option is B $\frac{4^n-2^n}{5^n}$

The given digits are even, which may be obtained as a product of two even numbers or a product of an even and an odd number.
Probability of obtaining $1,2,3,4,6,7,8,9$ at the end = ${\left(\frac{8}{10}\right)}^n={\left(\frac{4}{5}\right)}^n$.

Probability of obtaining $1,3,7,9$ at the end = ${\left(\frac{4}{10}\right)}^n={\left(\frac{2}{5}\right)}^n$.

Therefore, required probability = ${\left(\frac{4}{5}\right)}^n-{\left(\frac{2}{5}\right)}^n=\frac{4^n-2^n}{5^n}$