respectively. Using these formulae, the H(tL) and H(tR) for Mutation 1 is shown below: H(tL) = −(3/4log2(3/4) + 1/4log2(1/4)) = 0.811 H(tR) = −(1/3log2(1/3) + 2/3log2(2/3)) = 0.918 Following this, average entropy of the child nodes due to the split at node t of the above decision tree is computed as:H(s,t) = PLH(tL) + PRH(tR) where, probability of samples at the left child, PL = n(tL) / n(t), probability of samples at the right child, PR = n(tR) / n(t),Finally, H(s,t) along with PL and PR for Mutation 1 is as follows: PL = 4/7 PR = 3/7 H(s, t) = (4/70.811) + (3/70.918) = 0.857 Thus, by definition from equation (i):(Information gain) = H(t) - H(s,t) After all the steps, gain(s), where s is a candidate split for the example is: gain(s) = 0.985 – 0.857 = 0.128 Using this same set of formulae with the other three mutations leads to a table of the candidate splits, ranked by their information gain: The mutation that provides the most useful information would be Mutation 3, so that will be used to split the root node of the decision tree. The root can be split and all the samples can be passed though and appended to the child nodes. A tree describing the split is shown on the left. The samples that are on the left node of the tree would be classified as cancerous by the tree, while those on the right would be non-cancerous. This tree is relatively accurate at classifying the samples that were used to build it (which is a case of overfitting), but it would still classify sample C2 incorrectly. To remedy this, the tree can be split again at the child nodes to possibly achieve something even more accurate. To split the right node, information gain must again be calculated for all the possible candidate splits that were not used for previous nodes. So, the only options this time are Mutations 1, 2, and 4. Note: is different this time around since there are only four samples at the right child. PC, t = 1/4 PNC, t = 3/4 = -(1/4log2(1/4) + 3/4log2(3/4)) = 0.811 From this new , the candidate splits can be calculated using the same formulae as the root node: Thus, the right child will be split with Mutation 4. All the samples that have the mutation will be passed to the