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.

The chance that last digit of the product is $A_3$.

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

Answer 1

The correct option is D ${\left(\frac{4}{5}\right)}^n$

Let $n$ positive numbers be taken randomly and multiplied together.

$\Rightarrow A_3=$ last digit of the product among the no $:1,3,7or9$

$\Rightarrow A_2=$ last digit of the product among the no $:1,3,5,7or9$

$\Rightarrow$ Chance of selection of a no. among $A_3={\left(\frac{1}{4}\right)}^n$

$\Rightarrow$ Chance of selection of a no. in $A_2={\left(\frac{1}{5}\right)}^n$

$\therefore$ Chance that the last digit of product is $A_3=\frac{{\left(\frac{1}{5}\right)}^n}{{\left(\frac{1}{4}\right)}^n}$

$={\left(\frac{4}{5}\right)}^n.$

Hence, the answer is ${\left(\frac{4}{5}\right)}^n.$