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

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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}$

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