Question

In $n$ 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. $\frac{4^n}{5^n}$
• B. $\frac{8^n}{5^n}$
• C. $\frac{4^n-2^n}{5^n}$
• D. $\frac{8^n-4^n}{5^n}$

Answer 1

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

The last digits of the product will be $1,2,3,4,6,7,8$ and $9$ if and only if each of the n positive integers ends in any of three digits. now the probability of an integer ending in $1,2,3,4,6,7,8,9$ is $\frac{8}{10}=\frac{4}{5}$
Therefore the probability that the last digit of the product of n integers is $1,2,3,4,5,6,7,8$ or $9$ is ${\left(\frac{4}{5}\right)}^n$
Next, the last digits of the product will be $\mathrm{1.3.7}$ or $9$ if and if each of the n positive integers 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 probability of the required event is ${\left(\frac{4}{5}\right)}^n-{\left(\frac{2}{5}\right)}^n=\frac{4^n-2^n}{5^n}$