If a die is thrown twice,what is the probabil | vedclass

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(A) When a die is thrown twice,the total number of possible outcomes is $6 	imes 6 = 36$.Let $E$ be the event of getting $4$ at least once.It is easi

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$11/36$

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(A) When a die is thrown twice,the total number of possible outcomes is $6 	imes 6 = 36$.Let $E$ be the event of getting $4$ at least once.It is easier to find the probability of the complement event $E'$,which is the event of not getting $4$ in either throw.For a single throw,the outcomes that are not $4$ are $\{1, 2, 3, 5, 6\}$,which are $5$ outcomes.For two throws,the number of outcomes where $4$ does not appear is $5 	imes 5 = 25$.Therefore,the probability of not getting $4$ at all is $P(E') = \frac{25}{36}$.The probability of getting $4$ at least once is $P(E) = 1 - P(E') = 1 - \frac{25}{36} = \frac{11}{36}$.

If a die is thrown twice,what is the probability of getting $4$ at least once?

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