# Calculate probability using tree diagram and bayes theorem

We have the tree diagram as shown below,

I'm asked to find probability that a child who tests as right-handed will be left-handed?

I know we have to use Bayes Theorem and find ...

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First, let us apply Bayes formula as usual, then we will see if we can identify that as operations on the probability tree: $$\DeclareMathOperator{\P}{\mathbb{P}} \P(\text{L} \mid \text{test R}) =\frac{\P(\text{test R}\mid \text{L})\P(L)}{\P(\text{test R})}$$ Comparing this with the probability tree below, we see it involve all the nodes except the two "Tests as left-handed",
Then let us put the numbers into the Bayes formula above: $$\P(\text{L} \mid \text{test R}) =\frac{(0.1)\cdot(0)}{(0.1)\cdot (0) + (0.9)\cdot (0.95)} = 0$$ Then observe that in numerator we have the (sum of) path probabilities that passes through the node "Actually left-handed" (denoted L in the formulas here), while in the numerator we have the (sum of) all path probabilities that leads to one of the nodes "test as right-handed".
We can formulate that as a rule for applying Bayes theorem on a probability tree, for $\P(A \mid B)$, naming $A$ as 'cause' and $B$ as 'data':
1. Eliminate all the paths through the tree made impossible by the conditioning on data $B$.
2. Denominator is sum of all path-probabilities consistent with $B$
3. Numerator is sum of all path-probabilities consistent with cause $A$