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. $\frac{4^n-2^n}{5^n}$
• B. $\frac{2^n-2^n}{5^n}$
• C. $\frac{4^n-2^n}{3^n}$
• D. $\frac{4^{n-1}-2^n}{5^n}$

Answer 1

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

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