A fair die with faces numbered $1, ... , 6$ is thrown repeatedly. I want to figure out the probability that at least one 4 and at least one 5 occur before the first 6.
Before I start, I have looked at the mark-scheme and I understand the method they have used but with this question I am to understand where I am wrong.
My approach:
I want to figure out what is the complementary probability. The probability that there will be none $4$s and no $5$s before the first $6$ is
$$q_1= \frac{1}{6}+ \frac{3}{6}\frac{1}{6} + (\frac{3}{6})^2\frac{1}{6} + \cdots = \frac{1}{3}$$
The probability that there will be all $5$s (or similarly all $4$s) before the first 6 is
$$q_2=\frac{1}{6^2}+\frac{1}{6^3} + \cdots =\frac{1}{30}.$$
And thus the complimentary probability to the desired one is $$\frac{1}{3}+2 \times\frac{1}{30}=\frac{6}{15},$$
where the two comes from the fact that there is the scenario of all $4$s before a $6$ aswell. It follows that the desired probability is then
$$1-\frac{6}{15}=\frac{3}{5}.$$
However, the markscheme says he answer should be $\frac{1}{3}$. I assume I am wrong in my calculation of there beeing only of one type of number before the 6 but I don't see exactly where I am wrong. Could somebody assist me on this?
The complement event you want is that we either have zero $4$s or zero $5$s rolled before the $6$ is rolled. If we let $A$ be the event that no $4$s are rolled before the $6$ is rolled and $B$ be the event that no $5$s are rolled before the $6$ is rolled, then the probability we are after is $$1-P(A\cup B) = 1-P(A) - P(B) + P(A \cap B)$$
Can you take it from here?