📗 [4 points] John tells his professor that he forgot to submit his homework assignment. From experience, the professor knows that students who finish their homework on time forget to turn it in with probability . She also knows that of the students who have not finished their homework will tell her they forgot to turn it in. She thinks that of the students in this class completed their homework on time. What is the probability that John is telling the truth (i.e. he finished it given that he forgot to submit it)?

Hint

See Fall 2019 Final Q18 Q19, Fall 2017 Final Q6. Let \(C\) represent finishing (completing) homework and \(F\) represent forgetting to turn it. Then the question is asking \(\mathbb{P}\left\{C | F\right\} = \dfrac{\mathbb{P}\left\{C, F\right\}}{\mathbb{P}\left\{F\right\}} = \dfrac{\mathbb{P}\left\{F | C\right\} \mathbb{P}\left\{C\right\}}{\mathbb{P}\left\{F | C\right\} \mathbb{P}\left\{C\right\} + \mathbb{P}\left\{F | \neg C\right\} \left(1 - \mathbb{P}\left\{C\right\}\right)}\) due to the law of total probabilities.

📗 Answer: . Calculate

📗 [3 points] Assume the prior probability of having a female child (girl) is the same as having a male child (boy) and both are 0.5. The Smith family has kids. One day you saw one of the Smith children, and she is a girl. The Wood family has kids, too, and you heard that at least one of them is a girl. What is the chance that the Smith family has a boy? What is the chance that the Wood family has a boy?

Hint

See Fall 2014 Final Q10, Fall 2006 Final Q18, Fall 2005 Final Q18. In both cases, the probability is 1 minus the probability that all children are girls. The difference between the two cases is that for the Smith family with \(n_{S}\) kids, there are a total of \(2^{n_{S} - 1}\) possibilities (one of which is all of them are girls); and for the Wood family with \(n_{W}\) kids, there are a total of \(2^{n_{W}} - 1)\) possibilities (one of which is all of them are girls).

📗 Answer (comma separated vector): . Calculate

📗 [2 points] We have a biased coin with probability of producing Heads. We create a predictor as follows: generate a random number uniformly distributed in (0, 1). If the random number is less than we predict Heads, otherwise, we predict Tails. What is this predictor's (expected) accuracy in predicting the coin's outcome?

Hint

See Fall 2010 Final Q19. Suppose the probability of Heads is \(p\) and the probability of predicting Heads is \(q\), then the probability that the prediction is correctly Heads is \(p q\) and the probability that the prediction is correctly Tails is \(\left(1 - p\right)\left(1 - q\right)\). The accuracy is the sum of these two cases. By the way, the probability that a Head is predicted as Tail is \(p \left(1 - q\right)\) and the probability that a Tail is predicted as Head is \(q \left(1 - p\right)\). The sum of these four probabilities should be \(1\).

📗 Answer: . Calculate

📗 [4 points] Some Na'vi's don't wear underwear, but they are too embarrassed to admit that. A surveyor wants to estimate that fraction and comes up with the following less-embarrassing scheme: Upon being asked "do you wear your underwear", a Na'vi would flip a fair coin outside the sight of the surveyor. If the coin ends up head, the Na'vi agrees to say "Yes"; otherwise the Na'vi agrees to answer the question truthfully. On a very large population, the surveyor hears the answer "Yes" for fraction of the population. What is the estimated fraction of Na'vi's that don't wear underwear? Enter a fraction like 0.01 instead of a percentage 1%.

Hint

See Fall 2011 Midterm Q13. Let the fraction be \(f\), then the fraction of Yes would be \(0.5 + 0.5 \cdot \left(1 - f\right)\).

📗 Answer: . Calculate

📗 [4 points] Fill in the missing values in the following joint probability table so that A and B are independent.

Hint

See Fall 2019 Final Q20, Fall 2013 Final Q15, Fall 2011 Final Q4, Fall 2010 Final Q11.

📗 Answer (comma separated vector): . Calculate

📗 [3 points] There are two biased coins in my pocket: coin A has \(\mathbb{P}\left\{H | A\right\}\) = , coin B has \(\mathbb{P}\left\{H | B\right\}\) = . I took out a coin from the pocket at random with probability of A is . I flipped it twice the outcome is . What is the probability that the coin was ?

Hint

See Spring 2018 Final Q22 Q23, Fall 2018 Midterm Q11, Fall 2017 Final Q20, Spring 2017 Final Q6, Fall 2010 Final Q18. For example, the Bayes Rule for the probability that the document is \(A\) given the outcome is \(H T H\) is \(\mathbb{P}\left\{A | H T H\right\} = \dfrac{\mathbb{P}\left\{H T H, A\right\}}{\mathbb{P}\left\{H T H\right\}}\) = \(\dfrac{\mathbb{P}\left\{H T H | A\right\} \mathbb{P}\left\{A\right\}}{\mathbb{P}\left\{H T H | A\right\} \mathbb{P}\left\{A\right\} + \mathbb{P}\left\{H T H | B\right\} \mathbb{P}\left\{B\right\}}\) = \(\dfrac{\mathbb{P}\left\{H | A\right\} \mathbb{P}\left\{T | A\right\} \mathbb{P}\left\{H | A\right\} \mathbb{P}\left\{A\right\}}{\mathbb{P}\left\{H | A\right\} \mathbb{P}\left\{T | A\right\} \mathbb{P}\left\{H | A\right\} \mathbb{P}\left\{A\right\} + \mathbb{P}\left\{H | B\right\} \mathbb{P}\left\{T | B\right\} \mathbb{P}\left\{H | B\right\} \mathbb{P}\left\{B\right\}}\). Note that \(\mathbb{P}\left\{H T H | A\right\}\) can be split into three probabilities because the coins are independently flipped.

📗 Answer: . Calculate

📗 [3 points] You roll a 6-sided die times and observe the following counts in the table. Use Laplace smoothing (i.e. add-1 smoothing), estimate the probability of each side. Enter 6 numbers between 0 and 1, comma separated.

Hint

See Spring 2018 Final Q21, Fall 2016 Final Q4, Fall 2011 Midterm Q16. The maximum likelihood estimate of \(\mathbb{P}\left\{A = i\right\}\) is \(\dfrac{n_{i} + \delta}{\displaystyle\sum_{i'=1}^{6} n_{i'} + 6 \cdot \delta}\).

📗 Answer (comma separated vector): .  Calculate

📗 [2 points] In your day vacation, the counts of days are:

Use maximum likelihood estimate (no smoothing), estimate the probability that P(bighorn = | rainy = , warm = )?

Hint

See Fall 2017 Final Q3, Fall 2006 Final Q19, Fall 2005 Final Q19. For example, the maximum likelihood estimate of \(\mathbb{P}\left\{A | \neg B, C\right\} = \dfrac{\mathbb{P}\left\{A, \neg B, C\right\}}{\mathbb{P}\left\{\neg B, C\right\}}\), for binary variables \(A, B, C\), is \(\dfrac{n_{A, \neg B, C}}{n_{A, \neg B, C} + n_{\neg A, \neg B, C}}\).

📗 Answer: . Calculate

📗 [2 points] In a corpus with word tokens, the phrase "Fort Night" appeared times (not Fortnite). In particular, "Fort" appeared times and "Night" appeared . If we estimate probability by frequency (the maximum likelihood estimate) without smoothing, what is the estimated probability of P(Night | Fort)?

Hint

See Fall 2017 Midterm Q7, Fall 2016 Final Q4. The maximum likelihood estimate of \(\mathbb{P}\left\{B | A\right\} = \dfrac{\mathbb{P}\left\{A B\right\}}{\mathbb{P}\left\{A\right\}}\) is \(\dfrac{n_{A B}}{n_{A}}\).

📗 Answer: . Calculate

📗 [2 points] You have a vocabulary with \(n\) = word types. You want to estimate the unigram probability \(p_{w}\) for each word type \(w\) in the vocabulary. In your corpus the total word token count \(\displaystyle\sum_{w} c_{w}\) is , and \(c_{\text{dune}}\) = . Using Laplace smoothing (add ), compute \(p_{\text{dune}}\).

Hint

See Fall 2018 Midterm Q12. The maximum likelihood estimate of \(p_{w}\) is \(\dfrac{c_{w} + \delta}{\displaystyle\sum_{w'} c_{w'} + n \delta}\).

📗 Answer: . Calculate

📗 [1 points] Please enter any comments and suggestions including possible mistakes and bugs with the questions and the auto-grading, and materials relevant to solving the questions that you think are not covered well during the lectures. If you have no comments, please enter "None": do not leave it blank.
📗 Answer: .