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

Question

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

If there two events A = outcome 3 is first observed after 5 rolls and event B = outcome 3 is observed exactly three times in 10 rolls then what will be the probability.

Can I interpret this as P(3 observed 3 times in 10 rolls | 3 is first observed after 5 rolls)*P(3 is first observed after 5 rolls) / P(3 observed 3 times in 10 rolls)?

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

Answer 1

You need $$P (A \cap B)/P(B)$$ where $B$ is $3$ is observed $3$ times in $10$ rolls, and $A$ is $3$ appears first time on roll $5$. What you wrote is $$P(B|A) \cdot P(A)/P(B)$$ which equals the same thing.

We can calculate directly $$P(A\cap B) = (5/6)^4 \cdot (1/6) \cdot {5\choose 2}(1/6)^2 (5/6)^3$$ and $$P(B) = {10 \choose 3}(1/6)^3 (5/6)^7$$ which gives $$P(A\cap B)/P(B) = \frac{{5 \choose 2}}{{10 \choose 3}}=\frac{1}{12}$$

Note that even with a loaded die (or a tetrahedral one) you get the same result

Answer 2

For both outcomes we have to distribute 3 objects among 10 boxes. They do not need to be numbers. And it makes no difference what is their probability of selection compared with other objects, because these 3 objects have the same probability of being selected and it is already determined how many we are distributing.

In the 1st case there are no restrictions. Each can be placed in any box. In the 2nd case one object is placed in box 5 and the other two in boxes 6-10. We need to find the ratio of the number of distributions.

Distributing the other objects (digits) which are not 3s is the same in both cases, so we can ignore them. This will only introduce an extra factor in each case - the same factor - which will get cancelled in the ratio. Likewise the ratio is the same whether the objects are distinguishable or identical, so to simplify we can assume the 3 objects are distinguishable.

1st case : there are $10 \times 9 \times 8$ ways of placing the 3 objects among the 10 boxes.
2nd case : there are $3$ ways of placing an object in box 5 and $5 \times 4$ ways of placing the remaining 2 objects among boxes 6-10.

So the required probability is $\frac{3\times 5 \times 4}{10\times 9 \times 8}=\frac{1}{12}$.