'n' whole numbers are randomly chosen and multiplied:
Step-by-Step Solution
Key Concept: The probability of digit constraints in $n$ numbers involves counting favorable last digits and raising counts to the power $n$.
For part (A), the required event occurs when the last digit in all chosen numbers is from $\{1, 3, 7, 9\}$, so the probability is $(\frac{4}{10})^n$. For part (B), $P(\text{last digit is 1, 2, 3, 4, 6, 7, 8, 9}) - P(\text{last digit is 1, 3, 7, 9}) = \frac{8^n - 4^n}{10^n}$. For part (C), $P(1,3,5,7,9) - P(1,3,7,9) = \frac{5^n - 4^n}{10^n}$. For part (D), $P(0,5) - P(5) = \frac{(10^n - 8^n) - (5^n - 4^n)}{10^n}$.
Correct Answer: [A-r] [B-p][C-q][D-s]