Question

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) \(\frac{2^n}{5^n}\)

(b) \(\frac{4^n-2^n}{5^n}\)

(c) \(\frac{4^n}{5^n}\)

(d) None of these

Answer 1

(a) \(\frac{2^n}{5^n}\)

In any number the last digits can 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. Therefore, the last digit of each number can be chosen in 10 ways. 

∴ Exhaustive number of ways = 10ⁿ. If the last digit be 1, 3, 7 or 9, none of the numbers can be even or end in 0 or 5. That is, there is a choice of only 4 digits for the last digit of each of these n numbers. 

So, favourable number of ways = 4ⁿ 

∴ Required probability = \(\frac{4^n}{10^n}\) = \(\frac{2^n}{5^n}\).