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

Options

A. `(4^(n)+2^(n))/(5^(n))`
B. `(4^(n)xx2^(n))/(5^(n))`
C. `(4^(n)-2^(n))/(5^(n))`
D. None of these

Answer

C

Explanation

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 in 1, 2, 3, 4, 6, 7, 8 or 9 is `(8)/(10)=(4)/(5)`. Therefore, the probability that the last digit of the product of n integers ending in 1, 2, 3, 4, 6, 7, 8 or 9 is `((4)/(5))^(n)`. Next, the last digit of the product will be 1, 3, 7 or 9, if and only if each of the n positive integers ends in 1, 3, 7 or 9. The probability for an integer to end in 1, 3, 7 or 9 is `(4)/(10)=(2)/(5)` . Therefore, the probability for the product of n positive integers to end in 1, 3, 7 or 9 is `((2)/(5))^(n)`. Hence, the required probability
`=((4)/(5))^(n)-((2)/(5))^(n)=(4^(n)-2^(n))/(5^(n))`