Question

Let n positive integers taken randomly & multiplied together.Let $A_1$, $A_2$, $A_3$, $A_4$,$A_5$, $A_6$, $A_7$, be the events define as
$A_1$ = last digit of the product is not the number $1,3,7$ or$9$
$A_2$ = last digit of the product is among the number $1,3,5,7$ or $9$,
$A_3$ = last digit of the product is among the number $1,3,7$ or$9$
$A_4$ = last digit of lhe product 'is among the number $2,4,6$ or $8$.
$A_5$ = last digit of the product is the number $5$.
$A_6$ = last digit of the product is the number $0$
$A_7$ = last digit of the product is $1,2,3,4,6,7,8$ or$9$.On the basis of above information answer the following questions.

Then what is the probability that last digit of the product is $A_1$.?

• A. ${\left(\frac{1}{2}\right)}^n$
• B. ${\left(\frac{2}{5}\right)}^n$
• C. ${\left(\frac{3}{5}\right)}^n$
• D. ${\left(\frac{4}{5}\right)}^n$

Answer 1

Answer: (C)

${\left(\frac{3}{5}\right)}^n$

Now, A1= last digit in the product is not among the number $1,3,7$ or $9$ then it must be among $0,2,4,5,6$ or $8$.
It is possible only when digits taken are either $0$ or $2$ or $4$ or $5$or $6$ or $8$
$n\left(A_1\right)=6^{n}n\left(S\right)={10}^{n}P=\frac{6^n}{{10}^n}=\frac{3^n}{5^n}$