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
- the first two trials both result in an outcome of $1$
my attempt: let
-
{one $1st$} = event that outcome of first trial is one
-
{one $2nd$} = event that outcome of second trial is one
-
{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
$P(\text{one 1st}\cap \text{one 2nd}|\text{third last}) = \dfrac{P(\text{one 1st}\cap \text{one 2nd}\cap \text{third last})}{P(\text{third last})} =\dfrac{P(\text{one 1st})\cdot P(\text{one 2nd}|\text{one 1st})\cdot P(\text{third last}|\text{one 2nd}\cap \text{one 1st})}{P(\text{third last})} = \dfrac{\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{2}}{\frac{1}{3}} = \dfrac{1}{6}$
Best Answer
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.