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. $\frac{4" + 2"}{5"}$
• B. $\frac{4^{n}x2^{n}}{5^{n}}$
• C. $\frac{4^{n} - 2^{n}}{5^{n}}$
• D. None of these

Answer 1

Answer: (C)

$\frac{4^{n} - 2^{n}}{5^{n}}$

Answer 2

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 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 (4/5)ⁿ. 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 end in 1, 3, 7 or 9 is (2/5)ⁿ.

Hence the required probability = $\left( \frac{4}{5} \right)^{n} - \left( \frac{2}{5} \right)^{n} = \frac{4^{n} - 2^{n}}{5^{n}}$.