Suppose a fair six-sided die is rolled five times. Find the probability of 3 coming up at least once.

Question

Not really sure where to begin with this question. Obviously the probability of rolling a 3 is 1/6 for any given roll, but I don't really know how to extend this into this type of situation.

Answer 1

The complementary probability is no $3$ in all five trials which is

$$\left(1-\dfrac16\right)^5$$

Answer 2

To expound on lab bhattacharjee's answer, the easiest way to think about this is to think about the probability of rolling a 3 zero times, i.e., rolling anything other than a 3 five times in a row. Then subtract this from $1$.

Answer 3

To expand upon Y. Forman's expansion of lab bhattacharjee's answer, if you think about the probability of something happening, call it $P$, the chance of it not happening is $1-P$. So what is the chance of you never rolling a 3? That is easier to figure out. It's the chance of not rolling a 3 a three on any of the rolls, raised to the fifth power, as it has to happen five times. The chance of a three not being rolled, therefore, is $\left(5\over 6\right)^5$. This evaluates as $3125\over 7776$. The chance, then, of a three occurring at least once is $1$$-$$ {3125}\over 7776$, or $4651\over 7776$. In general, the chance of something happening at least once given $x$ identical chances to do so is $1-(1-P)^x$, where $P$ is the chance of it happening given one chance.