dataset_name: machine_learning
description: "The following are multiple choice questions (with answers) about machine\
  \ learning.\n\nQ: Which image data augmentation is most common for natural images?\n\
  (A) random crop and horizontal flip (B) random crop and vertical flip (C) posterization\
  \ (D) dithering\nA: Let's think step by step. Data augmentation is used to increase\
  \ the diversity of images in the training dataset. It is important that natural\
  \ images are kept natural after being augmented. Vertical flips of images are not\
  \ natural, so (B) is false. Posterization makes the image look like a poster and\
  \ and dithering increases color depth. None of these two preserve the natural property.\
  \ The only natural data augmentation technique is (A). The answer is (A).\n\nQ:\
  \ Traditionally, when we have a real-valued input attribute during decision-tree\
  \ learning we consider a binary split according to whether the attribute is above\
  \ or below some threshold. Pat suggests that instead we should just have a multiway\
  \ split with one branch for each of the distinct values of the attribute. From the\
  \ list below choose the single biggest problem with Pat\u2019s suggestion:\n(A)\
  \ It is too computationally expensive. (B) It would probably result in a decision\
  \ tree that scores badly on the training set and a testset. (C) It would probably\
  \ result in a decision tree that scores well on the training set but badly on a\
  \ testset. (D) It would probably result in a decision tree that scores well on a\
  \ testset but badly on a training set.\nA: Let's think step by step. Because the\
  \ input is real valued, it is unlikely that the same values appear both at training\
  \ and test time. This means that while such a decision tree could yield good performance\
  \ on the training data, when evaluated on the test data it will perform badly because\
  \ the decision tree won\u2019t know what to do with numbers that did not appear\
  \ in the training data. The answer is (C).\n\nQ: You are reviewing papers for the\
  \ World\u2019s Fanciest Machine Learning Conference, and you see submissions with\
  \ the following claims. Which ones would you consider accepting?\n(A) My method\
  \ achieves a training error lower than all previous methods! (B) My method achieves\
  \ a test error lower than all previous methods! (Footnote: When regularisation parameter\
  \ \u03BB is chosen so as to minimise test error.) (C) My method achieves a test\
  \ error lower than all previous methods! (Footnote: When regularisation parameter\
  \ \u03BB is chosen so as to minimise cross-validaton error.) (D) My method achieves\
  \ a cross-validation error lower than all previous methods! (Footnote: When regularisation\
  \ parameter \u03BB is chosen so as to minimise cross-validaton error.)\nA: Let's\
  \ think step by step. In machine learning, we train with some data and fixed hyperparameters\
  \ and the training error can be arbitrarily low, so (A) can\u2019t be right. Then,\
  \ one compares different hyperparameters by selecting the model with the lowest\
  \ cross-validation error, this means that (B) and (D) are not the right procedure.\
  \ The only relevant number after these is the test error and thus (C) is the right\
  \ answer. The answer is (C).\n\nQ: A 6-sided die is rolled 15 times and the results\
  \ are: side 1 comes up 0 times; side 2: 1 time; side 3: 2 times; side 4: 3 times;\
  \ side 5: 4 times; side 6: 5 times. Based on these results, what is the probability\
  \ of side 3 coming up when using Add-1 Smoothing?\n(A) 2.0/15 (B) 1.0/7 (C) 3.0/16\
  \ (D) 1.0/5\nA: Let's think step by step. Add-1 smoothing adds the value of one\
  \ to the different counts and then normalizes the probabilities accordingly. The\
  \ counts after adding one will be: side 1 comes up 1 time; side 2: 2 times; side\
  \ 3: 3 times; side 4: 4 times; side 5: 5 times; side 6: 6 times. The number of sum\
  \ one die rolls will be 21, so the probability of drawing a three is 3/21 = 1/7.\
  \ The answer is (B).\n\nQ: To achieve an 0/1 loss estimate that is less than 1 percent\
  \ of the true 0/1 loss (with probability 95%), according to Hoeffding's inequality\
  \ the IID test set must have how many examples?\n(A) around 10 examples (B) around\
  \ 100 examples (C) between 100 and 500 examples (D) more than 1000 examples\nA:\
  \ Let's think step by step. By the Hoeffding\u2019s inequality, we expect that with\
  \ 95% probability the in-sample and out-of-sample errors differ by epsilon when\
  \ we have N samples if 2 exp(-2 epsilon^2 N)<0.05, this implies that N > -1/(2*epsilon**2)\
  \ log ( 0.05/2 )= log (40)*5000. Since log(40)>1, we have that one needs more than\
  \ 1000 examples. The answer is (D)."
include: _mmlu_flan_cot_fewshot_template_yaml
task: mmlu_flan_cot_fewshot_machine_learning
