Independent rolls of a fair die

Question

Question: Let $X$ denote the number of independent rolls of a fair die required to obtain the first $3$. What is $P(X \ge 6)$?

I think I have to get $3$ in $X$ rolls and I think I have to get $x$-$1$ non $3$'s first followed by a $3$. Am I correct with the approach? If so, then how do I compute this because I am struggling to find a way to do it properly. Need help

Answer 1

The probability that $X\geq 6$ is simply the probability of rolling something other than $3$ on the first five rolls. Thus, $\left(\frac56\right)^5$.

Does that makes sense?

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Edit: You're right that $P(X=x)=(\frac56)^{x-1}(\frac16)$. If you sum that expression from $x=6$ to infinity, you end up with the same answer.

Answer 2

I think I have to get x-1 non 3's first followed by a 3.

Almost! You don't include that last 3 because that would mean $X = x$, but we are interested in $X\geq x$, where $x = 6$. Notice instead that $$\{X\geq6\}\iff\{ \text{The first five are not 3}\}.$$

This guarantees that the 3 will appear on the sixth roll or later.

Therefore $$P(X\geq 6) = P(\text{the first five are not 3}) = \left(\frac{5}{6}\right)^5.$$

Alternatively, notice that $X$ follows a geometric distribution with $p = 1/6$ on $\{1,2,3,\dotsc\}$.

Then we want $$P(X\geq 6) = P\left(\bigcup_{k = 6}^\infty X = k\right) = \sum_{k = 6}^\infty P(X=k)$$ where we can sum because the events are disjoint. You can verify that they are equal.

Answer 3

Expanding on G Tony Jacobs' answer.

I think I have to get $3$ in $X$ rolls and I think I have to get $x$-$1$ non $3$'s first followed by a $3$. Am I correct with the approach? If so, then how do I compute this because I am struggling to find a way to do it properly. Need help

Yes, $X$ is the count of tries until you obtain the first three. $X=x$ is the event of $x-1$ non-3 and then the first 3 on roll #$x$.

For the event of $X\geq 6$ you require non-3 in the first 5 rolls, then the first three on some roll after that. You don't care which. It could be the sixth, seventh, two thousand fourty second, or whatever; as long as it is not in the first five.

So $\mathsf P(X\geq 6)$ is the probability for obtaining five consecutive non-3 results, which is $(5/6)^5$.