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A fair coin with 1 marked on one face and 6 on the other and a fair dice are both tossed.Find the probability that the sum of the numbers that turn up is 3

$\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.
answered Jun 30, 2014 by sreemathi.v

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