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Consider an unending sequence of independent trials, where each trial is equally likely to result in any of the outcomes $1$, $2$, or $3$. Given that outcome $3$ is the last of the three outcomes to occur, find the conditional probability that

  1. the first two trials both result in an outcome of $1$

my attempt: let

  1. {one $1st$} = event that outcome of first trial is one

  2. {one $2nd$} = event that outcome of second trial is one

  3. {third last} = event that outcome three occurs after outcomes one and two have occurred.

$P(\text{one 1st}\cap \text{one 2nd}|\text{third last}) = \dfrac{P(\text{one 1st}\cap \text{one 2nd}\cap \text{third last})}{P(\text{third last})} = \dfrac{P(\text{third last}) \cdot P(\text{one 1st}|\text{third last}) \cdot P(\text{one 2nd}|\text{one 1st}\cap \text{third last})}{P(\text{third last})}$ $= P(\text{one 1st}|\text{third last}) \cdot P(\text{one 2nd}|\text{one 1st}\cap \text{third last})$

now, since each trial is equally likely to be either $1$, $2$, or $3$ and we are given that the $1^{st}$ trial is not $3$ hence, $P(\text{one 1st}|\text{third last})=0.5$

similarly, $P(\text{one 2nd}|\text{one 1st}\cap \text{third last}) = 0.5$ since all trials are independent, each trial is equally likely to be either $1$, $2$, or $3$ and the second trial's result cannot be $3$(since outcome $3$ occurs after outcomes $1$ and $2$ have both occurred)

hence, $P(\text{one 1st}\cap \text{one 2nd}|\text{third last}) =0.25$, but the given answer is $\dfrac{1}{6}.$

what did I do wrong?

edit: the given answer(which I understand) is
$P(\text{one 1st}\cap \text{one 2nd}|\text{third last}) = \dfrac{P(\text{one 1st}\cap \text{one 2nd}\cap \text{third last})}{P(\text{third last})} =\dfrac{P(\text{one 1st})\cdot P(\text{one 2nd}|\text{one 1st})\cdot P(\text{third last}|\text{one 2nd}\cap \text{one 1st})}{P(\text{third last})} = \dfrac{\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{2}}{\frac{1}{3}} = \dfrac{1}{6}$

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There are some assumption that is wrong.

For instance, the sequence $(1,1,3)$ is not a legitimate "third last" event but is counted as legitimate in your calculation.

The first "one first|third last" is correctly calculated to be $1\over 2$. However the "one second| one first and third last" is not $1\over 2$ because a $2$ has to occur somewhere after the $1$ and before $3$ so given the first is $1$ and the last is $3$, there is more chance that a $2$ occurs at the second trial.

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  • $\begingroup$ I finally understand what you mean by $(1,1,3)$ not being a legitimate event. The question says "last of the 3 outcomes". $\endgroup$
    – Ankit Seth
    Aug 26, 2021 at 3:29

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