When rolling a die, what is the probability that a 4 has appeared at least 3 times by the 15th roll?

Question

Now I've managed to figure out the probability that the 3rd time a 4 appears is on the 15th roll using a negative binomial distribution. But the at least is totally screwing with me in this variation of the problem.

Answer 1

Hint: the number of 4s in the first $15$ rolls is distributed $\mathrm{Bin}(15,1/6)$.

Answer 2

It is easier to compute the complementary probability

$$P(X\geq 3)=1-P(X=0)-P(X=1)-P(X=2)=$$

$$=1-\Bigg(\frac{5}{6}\Bigg)^{15}-15\cdot\frac{1}{6}\Bigg(\frac{5}{6}\Bigg)^{14}-\binom{15}{2}\Bigg(\frac{1}{6}\Bigg)^{2}\Bigg(\frac{5}{6}\Bigg)^{13}$$

Answer 3

'at least' means that 3 4s could appear anywhere in the 15 rolls, i.e. it is the probability that they appear in 3 rolls or in 4 rolls or in 5 rolls etc. up to 15 rolls. So it includes the possibility of getting a 4 on every roll.