If the four positive integers are selected randomly from the set of positive integers, then the probability that the number 1,3,7 and 9 are in the unit place in the product of 4−digit, so selected is
A. $\dfrac{7}{{625}}$
B. $\dfrac{2}{5}$
C. $\dfrac{5}{{625}}$
D. $\dfrac{{16}}{{625}}$

Hint: Since the product should end with either 1, 3, 7, or 9. So, we can say that numbers selected should not end with either of 0, 2, 4, 5, 6, 8 ( this can be checked like every digit from 0 to 9 is multiplied by 0, 2, 4, 5, 6, 8. And we notice that the unit digit is never 1,3,7 or 9). We can't take a number with unit digit 0, 2 , 4, 5, 6, 8.

Complete step by step solution:
When any number is multiplied with a number having unit digit 0, will always give a number having unit digit 0.
When any number is multiplied with a number having unit digit as even number then it will always give a number having unit digit as even number.
When any number is multiplied with a number having unit digit 5 then it'll always give a number having unit digit either 5 or 0.
So, we can take only those numbers which have unit digits either 1 or 3 or 7 or 9.
Now the sample space will be 10 since every number will have a unit digit ranging from 0 to 9. And the favorable cases are 4 (any number with unit digit 1, 3, 7, 9. (To select one number)
The product of any combination of these four numbers will always give a number with unit digit either 1 or 3 or 7 or 9.
Hence the probability will be ${\left( {\dfrac{4}{{10}}} \right)^4}$ = ${\left( {\dfrac{2}{5}} \right)^4}$ = $\dfrac{{16}}{{625}}$
Therefore, option (D) is the correct answer.

Note: All the 4 numbers should be odd To probability of randomly choosing an odd no. Is $\dfrac{1}{2}$. And also, the product should not end with digit five. So, the probability of choosing a multiple of 5 in this case is $\dfrac{1}{5}$.