Question

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

Answer 1

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:

$\frac{5}{6}\text{.}$

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

$\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\times\frac{5}{6}\text{.}$

Simplifying the above product we get:

$\frac{5^4}{6^4}=\frac{625}{1296}\text{.}$

Answer:

$\frac{625}{1296}\text{.}$