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# Two fair dice are tossed once. The probability of getting an even number on the first die or a total of 8 is

$\begin {array} {1 1} (1)\;\large\frac{4}{9} & \quad (2)\;\large\frac{5}{9} \\ (3)\;\large\frac{1}{3} & \quad (4)\;None\: of \: these \end {array}$

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## 1 Answer

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We have P ( even number on first die ) $= \large\frac{1}{12}$
P ( total of 8 ) = $\large\frac{5}{36}$
and P ( even number on first die and a total of 8 ) = $\large\frac{1}{12}$
Using addition theorem of probability we have P ( even number on
first die or a total of 8 )
$= \large\frac{1}{2} + \large\frac{5}{36}-\large\frac{1}{12}$
$= \large\frac{5}{9}$
Ans : (2)

answered Jan 2, 2014

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