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.
• B.
• C.
• D. None of these
  

Answer 1

The correct option is C

The last digit of the product will be 1, 2, 3, 4, 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, 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 ${\left(\frac{2}{5}\right)}^n$
Hence the required probability $={\left(\frac{4}{5}\right)}^n-{\left(\frac{2}{5}\right)}^n=\frac{4^n-2^n}{5^n}$