0
$\begingroup$

We have the tree diagram as shown below, enter image description here

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 ...

P(Actual Left handed | Tests as right-handed), however I'm not able to proceed further, Please help.

$\endgroup$
1
  • 1
    $\begingroup$ Welcome to Cross Validated! Please add the self-study tag and read its description. $\endgroup$ – DeltaIV Mar 3 '18 at 10:45
2
$\begingroup$

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",

enter image description here

so we can redraw the tree without those nodes:

enter image description here

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 $A$
$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ – metric Mar 4 '18 at 12:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.