# Find conditional of two normally distributed variables

Given two independent random variables (which describe two workers' durations to complete a task) $t_1, t_2$ which are normally distributed with ($\mu_1=20,\sigma_1=5$) and ($\mu_2=30,\sigma_2=10$). A task is given at random to worker 1 or 2 with equal probability of $p=0.5$.

A task is completed in $t=10$. Now, I would like to find the probability of this task being completed by worker 2.

It would be great if someone could guide me through this task. So far I came up with the following thought-process:

1) I assume workers durations to be independent, thus $p(t)=p(t_1)p(t_2)$ (But not sure what to do with the equal probability of 0.5)

2) My goal is to find $p(\text{worker}_2\vert t=10)$ but I am not sure how?

• Use $p(t)=\mathcal{N}(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2 )$
There are two type of workers, $W = 1$ and $W=2$. Given an observation of completion time $t$, we can use Bayes' rule to compute the probability that this observation was produced by a worker of type $i$: $$p(W=i|t) = \frac{p(t|W=i)\,p(W=i)}{p(t)} ,$$ where $$p(t) = \sum_{i=1}^2 p(t|W=i)\,p(W=i) .$$
In the setup we are told $p(W=1) = p(W=2) = 1/2$, $p(t|W=1) = \textsf{N}(t|20,5^2)$, and $p(t|W=2) = \textsf{N}(t|30,10^2)$. Therefore, $$p(W=2|t=10) = \frac{\textsf{N}(10|30,10^2)\times \frac{1}{2}}{\textsf{N}(10|20,5^2)\times \frac{1}{2} + \textsf{N}(10|30,10^2)\times \frac{1}{2}} = \frac{1}{3} .$$