The chance of throwing a total of $3$ or $5$ or $11$ with two dice is:

\frac{5}{36}

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\frac{1}{9}

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\frac{2}{9}

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\frac{19}{36}

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Solution

The correct option is C $\frac{2}{9}$
The number of possible combinations with two dice is $6 \times 6 = 36$ .
The number of ways of getting $3$ as sum is $2$ .
The number of ways of getting $5$ as sum is $4$ .
The number of ways of getting $11$ as sum is $2$ .
The number of ways of getting $3$ or $5$ or $11$ as sum is $8$ .
The probability is $\frac{8}{36} = \frac{2}{9}$ .
Therefore option $C$ is correct.

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Pls solve these question with steps

=or > or <

(-3)+7-(19) 15-8+(-9)

23-41+11 . 15-8+(-9)

39+(-24)-(15) . 36+(-52)-(-36)

-231 +79+51 . -399+159+81

Pls solve with step

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