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

\frac{1}{36}

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\frac{3}{36}

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\frac{11}{36}

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\frac{20}{36}

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Solution

The correct option is D $\frac{20}{36}$
$A = getting even no on 1st dice$
$B = getting sum 8$
So, $n (A) = 18$
So, $P (A \cup B) = \frac{18 + 5 - 3}{36} = \frac{20}{36}$

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

The correct option is D 2036A=getting even no on 1st diceB=getting sum 8So, n(A)=18So, P(A∪B)=18+5−336=2036

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

The correct option is D $\frac{20}{36}$
$A = getting even no on 1st dice$
$B = getting sum 8$
So, $n (A) = 18$
So, $P (A \cup B) = \frac{18 + 5 - 3}{36} = \frac{20}{36}$