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A fair die is rolled four times. What is the probability of obtaining no 6's?

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To solve this question we will use the following expression to compute the theoretical probability:


\frac{\text{favorable cases}}{total\text{ cases}}\text{.}

Now, recall that the possible outcomes of rolling a die are 1, 2, 3, 4, 5, and 6. Therefore the probability of obtaining no 6 in each rolling is:


(5)/(6)\text{.}

Then, the probability of obtaining no 6's in four rollings is:


(5)/(6)*(5)/(6)*(5)/(6)*(5)/(6)\text{.}

Simplifying the above product we get:


(5^4)/(6^4)=(625)/(1296)\text{.}

Answer:


(625)/(1296)\text{.}

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