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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.

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  • $\begingroup$ Don't you mean $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$? $\endgroup$
    – Toni
    Apr 10, 2020 at 11:23
  • $\begingroup$ "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). $\endgroup$ Apr 10, 2020 at 11:42
  • $\begingroup$ @Toni Yes! but isn't B here is a subset of A. I have edited the assumption I used in my question. $\endgroup$ Apr 10, 2020 at 12:31
  • $\begingroup$ @BarryCipra we get the first 3 on the 5th roll. $\endgroup$ Apr 10, 2020 at 12:31
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    $\begingroup$ But in your example $B$ is not a subset of $A$. $\endgroup$
    – Toni
    Apr 10, 2020 at 12:58

2 Answers 2

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I interpret your question as follows: We want to find the conditional probability that in total there are exactly three 3's rolled given that the first four rolls are not equal 3, but the fifth is equal 3.

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, $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{(3/4)^4(1/4) {5\choose 2}(1/4)^2(3/4)^{5-2}}{(3/4)^4(1/4)}=10(1/4)^2(3/4)^{3}=\frac{270}{4^5}=\frac{135}{512}.$$

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If the first "3" occurs on the fifth roll, then those first five rolls can be ignored. (The question said this was "given", so it's already happened, with 100% certainty.)

Knowing that, we can reduce the question to finding the probability of rolling exactly two "3"s in the next five rolls.

And that much simpler situation, you should be able to analyze quite easily.

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