$\begin{array}{1 1}(A)\;\large\frac{1}{12}\\(B)\;\large\frac{3}{22}\\(C)\;\large\frac{5}{12}\\(D)\;\text{None of these}\end{array} $

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Step 1:

Given :

Two sides of a coin marked as 1 & 6

A die is also tossed

When both are tossed,the sample space (or) the total number of outcomes

$\Rightarrow 2\times 6=12$

Step 2:

Probability that sum of numbers turns up as 3

$\therefore$ The sum can turn up to be 3 only when 1 appears on the coin and 2 appears on the die.

$\therefore$ Only possibility =(1,2)

Step 3:

Required probability =$\large\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

$\Rightarrow \large\frac{1}{12}$

Hence (A) is the correct answer.

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