# 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 ...

• Welcome to Cross Validated! Please add the self-study tag and read its description. Mar 3, 2018 at 10:45

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", so we can redraw the tree without those nodes: 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 $A$
• Concept is now more clear, however we want to calculate probability that a child who test right will be a left handed. Also I understand there is a typo and writing numerator twice. Mar 4, 2018 at 12:11