# What is the probability that the outcome 3 is observed exactly three times in 10 rolls given that it is first observed after 5 rolls?

What is the probability that the outcome 3 is observed exactly three times in 10 rolls given that it is first observed after 5 rolls? using a tetrahedron dice.

If there two events $$A$$ = outcome $$3$$ is observed exactly three times in 10 rolls and event $$B$$ = first observed after 5 rolls then what will be the probability. A tetrahedron dice has 4 faces. Considering the scenario given I assume it is :

When $$B$$ is a subset of $$A$$: If $$B\subset A$$, then whenever $$B$$ happens, $$A$$ also happens. Thus, given that $$B$$ occurred, we expect that the probability of $$A$$ be one. In this case $$A\cap B=B$$, so $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B)}{P(B)}=1.$$

Is my assumption correct?

First, observe after 5th roll indicates that the first 3 we get is on 5th roll.

• Don't you mean $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$?
– Toni
Apr 10, 2020 at 11:23
• "after $5$ rolls" is somewhat ambiguous: Does it mean no sooner than the sixth roll but possibly later, or does it mean on the sixth roll (but no sooner). Apr 10, 2020 at 11:42
• @Toni Yes! but isn't B here is a subset of A. I have edited the assumption I used in my question. Apr 10, 2020 at 12:31
• @BarryCipra we get the first 3 on the 5th roll. Apr 10, 2020 at 12:31
• But in your example $B$ is not a subset of $A$.
– Toni
Apr 10, 2020 at 12:58

To stick with your notation, let $$A$$ be the event that there are exactly three 3's in total and $$B$$ the event that the first 4 rolls are not 3's and the fifth is a 3. Then, $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{(3/4)^4(1/4) {5\choose 2}(1/4)^2(3/4)^{5-2}}{(3/4)^4(1/4)}=10(1/4)^2(3/4)^{3}=\frac{270}{4^5}=\frac{135}{512}.$$