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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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A)
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)

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A)
total number of outcomes are 6*6=36 condition1: let P(A) be the probability of getting an even number on the first die, first die can have 2,4,6 and 2nd die can have any of 1,2,3,4,5,6 favorable outcomes=3*6=18 therefore P(A)=18/36=1/2 condition 2: let P(B) be the probability of getting total of 8. therefore favorable outcomes={(2,6),(3,5),(4,4),(5,3),(6,2)} P(B)=5/36 now both the above cases have some cases common, i.e. when we are getting total of 8 and first die has even number. the common cases are {(2,6),(4,4),(6,2)} therefore P(A∩B)=3/36=1/12 hence the required probability: P(A∪B)=P(A)+P(B)-P(A∩B) P(even on first die or total of 8)=1/2+5/36-1/12=5/9