diff --git a/source/main.ptx b/source/main.ptx index cdac259..fa994b5 100644 --- a/source/main.ptx +++ b/source/main.ptx @@ -59,18 +59,18 @@ - + --> Exam Review - - + - + - -

- In each of the following scenarios with given events A and B, alculate \Pr(A), \Pr(B), \Pr(A\cap B), \Pr(A \mid B), and \Pr(B \mid A). -

- - - - -

- An experiment consists of rolling a fair die two times. - Let A be the event that the sum is even, and let B be the event that the second roll is higher than the first. -

- - - -

- - A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), - \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), - \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} - B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), - \amp (2, 3), (2, 4), (2, 5), (2, 6), - \amp (3, 4), (3, 5), (3, 6), - \amp (4, 5), (4, 6), - \amp (5, 6)\} - A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} - - So \Pr(A) = \frac{18}{36} = \frac{1}{2}, \Pr(B) = \frac{15}{36} = \frac{5}{12}, and \Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}. - Finally: - - \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} - \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{6/36}{18/36} = \frac{6}{18} = \frac{1}{3} - -

- - - - - -

- An experiment consists of flipping a fair coin three times. - Let A be the event that the first and second flips match. - Let B be the event that there are at least two heads. -

- - - -

- - A \amp = \{ HHH, HHT, TTH, TTT \} - B \amp = \{ HHH, HHT, HTH, THH \} - A\cap B \amp = \{HHH, HHT\} - - So \Pr(A) = \frac{4}{8} = \frac{1}{2}, \Pr(B) = \frac{4}{8} = \frac{1}{2}, and \Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}. - Finally: - - \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} - \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} - -

- - - - - - -

- A diagnostic test is developed to detect a disease present in 3.2% of the population. - For a patient who has the disease, the test will accurately give a positive result 65% of the time. - When the patient does not have the disease, the test will accurately give a negative result 99.9% of the time. -

- - - - - -

- For a patient who receives a positive test, what is the probability they have the disease? -

- - - -

- Let P be the event of testing positive and D the event of having the disease. - Then the prevalence \Pr(D) is given as 3.2%, or 0.032. - The sensitivity is \Pr(P\mid D) = 0.65, and the specificity is \Pr(P^c\mid D^c) = 0.999. - So, according to Bayes' Theorem: - - \Pr(D\mid P) \amp = \frac{\Pr(P\mid D)\Pr(D)}{\Pr(P\mid D)\Pr(D) + (1 - \Pr(P^c\mid D^c))\Pr(D^c)} - \amp = \frac{(0.65)(0.032)}{(0.65)(0.032) + (1 - 0.999)(1 - 0.032)} - \amp \approx 0.96 - -

- - - - - - -

- For a patient who receives a negative test, what is the probability they do not have the disease? + Suppose we have a 6-sided die that's weighted to roll a 6 half of the + time. + We roll the die two times. + List the set of all possible results. + [Note: the result (2, 4)---rolling a 2 and then a 4---is different + from the result (4, 2)---rolling a 4 and then a 2.]

- \Pr(D^c\mid P^c) \amp = \frac{\Pr(P^c\mid D^c)\Pr(D^c)}{\Pr(P^c\mid D^c)\Pr(D^c) + (1 - \Pr(P\mid D))\Pr(D)} - \amp = \frac{(0.999)(1 - 0.032)}{(0.999)(1 - 0.032) + (1 - 0.65)(0.032)} - \amp \approx 0.99 + \Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), + \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), + \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), + \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), + \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), + \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}

- - - - -

- An experiment consists of rolling a fair die two times. - Let A be the event that the sum is even, and let B be the event that the second roll is higher than the first. - Are A and B independent? -

- + +

+ + \Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), + \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), + \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), + \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), + \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), + \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} + + Note that \Omega simply lists outcomes with no reference to the + probabilities. +

+ + - -

- - A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), - \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), - \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} - B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), - \amp (2, 3), (2, 4), (2, 5), (2, 6), - \amp (3, 4), (3, 5), (3, 6), - \amp (4, 5), (4, 6), - \amp (5, 6)\} - A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} - - So \Pr(A) = \frac{18}{36} = \frac{1}{2}, \Pr(B) = \frac{15}{36} = \frac{5}{12}, and \Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}. - Finally: - - \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \neq \Pr(A), - - so A and B are not independent. -

- - - - - -

- An experiment consists of flipping a fair coin three times. - Let A be the event that the first and second flips match. - Let B be the event that there are at least two heads. - Are A and B independent? -

- - - -

- - A \amp = \{ HHH, HHT, TTH, TTT \} - B \amp = \{ HHH, HHT, HTH, THH \} - A\cap B \amp = \{HHH, HHT\} - - So \Pr(A) = \frac{4}{8} = \frac{1}{2}, \Pr(B) = \frac{4}{8} = \frac{1}{2}, and \Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}. - Finally: - - \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} = \Pr(A), - - so A and B are independent. -

- - - - - -

- Let A = \{1, 2, 3\} and B = \{3, 4, 5\} be events in the sample space \Omega = \{1, 2, 3, 4, 5, 6\}. - Create a probability distribution for \Omega so that A, B are independent. -

- - - - - Example Distribution - - - - x - \Pr(x) - - - - 1 - 0.1 - - - - 2 - 0.2 - - - - 3 - 0.2 - - - - 4 - 0.1 - - - - 5 - 0.1 - - - - 6 - 0.3 - - -
- -

- Now \Pr(A) = 0.5, \Pr(B) = 0.4, and - - \Pr(A\cap B) = 0.2 = (0.5)(0.4) = \Pr(A)\Pr(B), - - so A and B are independent. -

- - - - - -

- An experiment consists of flipping a biased coin 20 times. - If the coin comes up heads with probability p = 0.3, find the probability of seeing 5 heads. - Find the probability of seeing up to (and including) 3 heads. -

- - - -

- Let S be the number of heads. - Then S \sim \Bin(20, 0.3), so: - - \Pr(S = 5) \amp = {20 \choose 5} (0.3)^5 (0.7)^{20 - 5} - \amp = \frac{20!}{(5!)(15!)} (0.3)^5 (0.7)^{15} - \amp = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} (0.3)^5 (0.7)^{15} - \amp = (19 \times 3 \times 17 \times 16) (0.3)^5 (0.7)^{15} - \amp \approx 0.179 - -

- - - - - -

- An experiment consists of flipping a coin repeatedly until we first see heads. -

- - - - +

- If the coin comes up heads with probability 0.4, what is the probability we'll see our first heads within three flips? What about precisely on the third flip? + Suppose we flip a coin two times. + List the set of all possible results. + What about flipping three times? + Four times? + If we flip the coin 10 times, how many possible results will there be? +

+ + + +

+ For two flips: \Omega = \{ HH, HT, TH, TT \}. +

+ +

+ For three flips: + + \Omega = \{ \amp HHH, HHT, HTH, THH, + \amp HTT, THT, TTH, TTT \} + +

+ +

+ For four flips: + + \Omega = \{ \amp HHHH, HHHT, HHTH, HTHH, + \amp THHH, HHTT, HTHT, HTTH, + \amp THHT, THTH, TTHH, HTTT, + \amp THTT, TTHT, TTTH, TTTT \}. + +

+ +

+ Each additional flip doubles the number of outcomes. + So, with ten flips, we'll have |\Omega| = 2^{10} = 1024. +

+ + + + + +

+ If we roll a 6-sided die ten times, how many possible results will + there be? +

+ + + +

+ |\Omega| = 6^{10}. +

+ + + +

+ Each additional roll will multiply the number of outcomes by 6. + So, with 10 rolls, we'll have |\Omega| = 6^{10}. +

+ + + + + +

+ Consider the sample space \Omega = \{1, 2, 3, 4, 5, 6, 7, 8\} + with probability distribution below. + Calculate the probabilities of A = \{1, 3, 7, 8\}, + B = \{2, 3, 6, 7\}, A\cup B, and A \cap B. +

+ + + + <tabular halign="center"> + <row bottom="minor"> + <cell><m>x</m></cell> + <cell><m>\Pr(x)</m></cell> + </row> + + <row> + <cell>1</cell> + <cell>0.1</cell> + </row> + + <row> + <cell>2</cell> + <cell>0.05</cell> + </row> + + <row> + <cell>3</cell> + <cell>0.2</cell> + </row> + + <row> + <cell>4</cell> + <cell>0.15</cell> + </row> + + <row> + <cell>5</cell> + <cell>0.15</cell> + </row> + + <row> + <cell>6</cell> + <cell>0.1</cell> + </row> + + <row> + <cell>7</cell> + <cell>0.05</cell> + </row> + + <row> + <cell>8</cell> + <cell>0.1</cell> + </row> + + <row> + <cell>9</cell> + <cell>0.1</cell> + </row> + </tabular> + </table> + </statement> + + <hint> + <p> + Remember that, to calculate the probability of an event, you should + add up the probabilities of each outcome in the event. + </p> + </hint> + + <answer> + <p> + <m>\Pr(A) = 0.45, \Pr(B) = 0.4, \Pr(A \cup B) = 0.6, \Pr(A \cap B) = 0.25.</m> + </p> + </answer> + + <solution> + <p> + <md> + <mrow> \Pr(A) \amp = \Pr(1) + \Pr(3) + \Pr(7) + \Pr(8) </mrow> + <mrow> \amp = 0.1 + 0.2 + 0.05 + 0.1 </mrow> + <mrow> \amp = 0.45 </mrow> + <mrow> \Pr(B) \amp = \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) </mrow> + <mrow> \amp = 0.05 + 0.2 + 0.1 + 0.05 </mrow> + <mrow> \amp = 0.4 </mrow> + </md> + The other events are <m>A\cup B = \{1, 2, 3, 6, 7, 8\}</m> and + <m>A\cap B = \{3, 7\}</m>, so: + <md> + <mrow> \Pr(A \cup B) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(6) + \Pr(7) + \Pr(8) </mrow> + <mrow> \amp = 0.1 + 0.05 + 0.2 + 0.1 + 0.05 + 0.1 </mrow> + <mrow> \amp = 0.6 </mrow> + <mrow> \Pr(A \cap B) \amp = \Pr(3) + \Pr(7) </mrow> + <mrow> \amp = 0.2 + 0.05 </mrow> + <mrow> \amp = 0.25 </mrow> + </md> + </p> + </solution> + </exercise> + + <exercise> + <introduction> + <p> + Suppose we flip a coin two times. + Answer the questions below. + What about three flips? + What about four flips? + </p> + </introduction> + + <task> + <statement> + <p> + Write all outcomes in the sample space <m>\Omega</m>. + </p> + </statement> + + <answer> + <p> + <m>\Omega = \{HH, HT, TH, TT\}.</m> + </p> + </answer> + </task> + + <task> + <statement> + <p> + Make a probability distribution table for <m>\Omega</m> assuming the + coin is fair. + </p> + </statement> + + <answer> + <table> + <title>Probability Distribution for Two Fair Coin Flips + + + + x + \Pr(x) + + + + HH + 0.25 + + + + HT + 0.25 + + + + TH + 0.25 + + + + TT + 0.25 + + +
+ + + + + +

+ Make a probability distribution table assuming the coin comes up + heads with probability 0.3. +

+ + + + + Probability Distribution for Two Fair Coin Flips + + + + x + \Pr(x) + + + + HH + 0.09 + + + + HT + 0.21 + + + + TH + 0.21 + + + + TT + 0.49 + + +
+ + + + + + +

+ Suppose we roll a fair 6-sided die two times. + Answer the questions below. +

+ + + + +

+ Write all outcomes in the sample space \Omega. +

+ + + +

+ + \Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), + \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), + \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), + \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), + \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), + \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} + +

+ + + + + +

+ Make a probability distribution table for \Omega assuming the + die is fair. +

+ + + +

+ We'll avoid an overly large table and note that, since the die is + fair, every outcome is equally likely. + Therefore, \Pr(x) = \frac{1}{36} for every + x\in \Omega. +

+ + + + + +

+ Let A be the event that the second roll is higher than the + first, and let B be the event that the first roll is even. + Find \Pr(A), \Pr(B), and \Pr(A \cap B). +

+ + + +

+ \Pr(A) = \frac{5}{12}, \Pr(B) = \frac{1}{2}, \Pr(A \cap B) = \frac{1}{6}. +

+ + + +

+ + A = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), + \amp (2, 3), (2, 4), (2, 5), (2, 6), + \amp (3, 4), (3, 5), (3, 6), + \amp (4, 5), (4, 6), + \amp (5, 6)\} + B = \{ \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), + \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), + \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} + A \cap B = \{ \amp (2, 3), (2, 4), (2, 5), (2, 6), + \amp (4, 5), (4, 6)\} + + Therefore + \Pr(A) = \frac{15}{36} = \frac{5}{12}, \Pr(B) = \frac{18}{36} = \frac{1}{2}, \Pr(A \cap B) = \frac{6}{36} = \frac{1}{6}. +

+ + + + + + +

+ Suppose a die has the values 1, 2, 3, 4, 5, 6 on the faces, but + the die is not fair. + Instead, the probabilities scale by the same amount as the face + values. + For example, a result of 4 is twice as likely as a result of 2, since + 4 is twice as large as 2; a result of 6 is six times more likely than + a result of 1; and so on. + Write a probability distribution table for this die. +

+ + + + + Probability Distribution for a Linearly Scaled Die + + + + x + \Pr(x) + + + + 1 + 1/21 + + + + 2 + 2/21 + + + + 3 + 3/21 + + + + 4 + 4/21 + + + + 5 + 5/21 + + + + 6 + 6/21 + + +
+ + + +

+ Let \Pr(1) = x. + Then \Pr(2) = 2 \Pr(1) = 2x, and \Pr(3) = 3\Pr(1) = 3x, + and so on. + So the total probability in the space is: + + Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) + \amp = x + 2x + 3x + 4x + 5x + 6x + \amp = 21x + + Since the total probability must add up to 1, we have 1 = 21x, + so x = \frac{1}{21}, and we can calculate the rest of the + probabilities from there. +

+ + + Probability Distribution for a Linearly Scaled Die + + + + x + \Pr(x) + + + + 1 + 1/21 + + + + 2 + 2/21 + + + + 3 + 3/21 + + + + 4 + 4/21 + + + + 5 + 5/21 + + + + 6 + 6/21 + + +
+ + + + + +

+ Suppose a die has the values 1, 2, 3, 4, 5, 6 on the faces, but + the die is not fair. + Instead, each even value has an equal probability, each odd value has + an equal probability, and the even values are each twice as likely as + the odd values to appear on a roll. + Write a probability distribution table for this die. +

+ + + + + Probability Distribution for an Even-biased Die + + + + x + \Pr(x) + + + + 1 + 1/9 + + + + 2 + 2/9 + + + + 3 + 1/9 + + + + 4 + 2/9 + + + + 5 + 1/9 + + + + 6 + 2/9 + + +
+ + + +

+ Let \Pr(1) = x. + Then \Pr(3) = x and \Pr(5) = x, since all odd rolls must + have the same probability. + The even rolls must have twice the probability, so + \Pr(2) = \Pr(4) = \Pr(6) = 2x. + Now: + + Pr(\Omega) \amp = \Pr(1) + \Pr(2) + \Pr(3) + \Pr(4) + \Pr(5) + \Pr(6) + \amp = x + 2x + x + 2x + x + 2x + \amp = 9x + + Since the total probability must add up to 1, we have 1 = 9x, + so x = \frac{1}{9}, and we can calculate the rest of the + probabilities from there. +

+ + + Probability Distribution for an Even-biased Die + + + + x + \Pr(x) + + + + 1 + 1/9 + + + + 2 + 2/9 + + + + 3 + 1/9 + + + + 4 + 2/9 + + + + 5 + 1/9 + + + + 6 + 2/9 + + +
+ + + + + +

+ A toxin molecule inside a cell has a 0.3 probability of leaving the + cell during a 1-minute period. + For each value of n = 1, 2, 3, \dotsc, find the probability of + the toxin molecule leaving the cell during the n th minute. + What is the probability of the molecule leaving the cell during the + first 3 minutes? +

+ + + +

+ Write T for the minute that the toxin molecule leaves the cell. + Then \Pr(T = n) = (0.7)^{n - 1} (0.3), and + \Pr(T \leq 3) = 0.657. +

+ + + +

+ Write T for the minute that the toxin molecule leaves the cell. + We're told that the toxin molecule has probability 0.3 of + leaving during each minute. + Therefore, the probability of remaining during a particular minute is + 1 - 0.3 = 0.7. +

+ +

+ For the toxin molecule to leave the cell during minute n, it + must remain for minutes 1, 2, \dotsc, n - 1, and then leave + during minute n. + The probability to remain each minute is 0.7, so we must + multiply n - 1 copies of 0.7. + Then the probability of leaving is 0.3, so we multiply by a + factor of 0.3, yielding the formula: + + \Pr(T = n) = (0.7)^{n - 1}(0.3). + +

+ +

+ Then, the probaiblity of the molecule leaving within the first three + minutes will be: + + \Pr(T \leq 3) \amp = \Pr(T = 1) + \Pr(T = 2) + \Pr(T = 3) + \amp = 0.3 + (0.7)(0.3) + (0.7)^2(0.3) + \amp = 0.657. + +

+ + + + +

+ In each of the following scenarios with given events A and + B, calculate \Pr(A), \Pr(B), \Pr(A\cap B), + \Pr(A \mid B), and \Pr(B \mid A). +

+ + + + +

+ An experiment consists of rolling a fair die two times. + Let A be the event that the sum is even, and let B be + the event that the second roll is higher than the first. +

+ + + +

+ + A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), + \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), + \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} + B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), + \amp (2, 3), (2, 4), (2, 5), (2, 6), + \amp (3, 4), (3, 5), (3, 6), + \amp (4, 5), (4, 6), + \amp (5, 6)\} + A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} + + So \Pr(A) = \frac{18}{36} = \frac{1}{2}, + \Pr(B) = \frac{15}{36} = \frac{5}{12}, and + \Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}. + Finally: + + \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} + \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{6/36}{18/36} = \frac{6}{18} = \frac{1}{3} + +

+ + + + + +

+ An experiment consists of flipping a fair coin three times. + Let A be the event that the first and second flips match. + Let B be the event that there are at least two heads. +

+ + + +

+ + A \amp = \{ HHH, HHT, TTH, TTT \} + B \amp = \{ HHH, HHT, HTH, THH \} + A\cap B \amp = \{HHH, HHT\} + + So \Pr(A) = \frac{4}{8} = \frac{1}{2}, + \Pr(B) = \frac{4}{8} = \frac{1}{2}, and + \Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}. + Finally: + + \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} + \Pr(B \mid A) \amp = \frac{\Pr(B\cap A)}{\Pr(A)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} + +

+ + + + + + +

+ A diagnostic test is developed to detect a disease present in 3.2% of + the population. + For a patient who has the disease, the test will accurately give a + positive result 65% of the time. + When the patient does not have the disease, the test will accurately + give a negative result 99.9% of the time. +

+ + + + +

+ For a patient who receives a positive test, what is the probability + they have the disease? +

+ + + +

+ \approx 0.96. +

+ + + +

+ Let P be the event of testing positive and D the event + of having the disease. + Then the prevalence \Pr(D) is given as 3.2%, or 0.032. + The sensitivity is \Pr(P\mid D) = 0.65, and the specificity + is \Pr(P^c\mid D^c) = 0.999. + So, according to Bayes' Theorem: + + \Pr(D\mid P) \amp = \frac{\Pr(P\mid D)\Pr(D)}{\Pr(P\mid D)\Pr(D) + (1 - \Pr(P^c\mid D^c))\Pr(D^c)} + \amp = \frac{(0.65)(0.032)}{(0.65)(0.032) + (1 - 0.999)(1 - 0.032)} + \amp \approx 0.96 + +

+ + + + + +

+ For a patient who receives a negative test, what is the probability + they do not have the disease? +

+ + + +

+ \approx 0.99. +

+ + + +

+ + \Pr(D^c\mid P^c) \amp = \frac{\Pr(P^c\mid D^c)\Pr(D^c)}{\Pr(P^c\mid D^c)\Pr(D^c) + (1 - \Pr(P\mid D))\Pr(D)} + \amp = \frac{(0.999)(1 - 0.032)}{(0.999)(1 - 0.032) + (1 - 0.65)(0.032)} + \amp \approx 0.99 + +

+ + + + + + +

+ An experiment consists of rolling a fair die two times. + Let A be the event that the sum is even, and let B be + the event that the second roll is higher than the first. + Are A and B independent? +

+ + + +

+ A and B are not independent. +

+ + + +

+ + A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), + \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), + \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\} + B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), + \amp (2, 3), (2, 4), (2, 5), (2, 6), + \amp (3, 4), (3, 5), (3, 6), + \amp (4, 5), (4, 6), + \amp (5, 6)\} + A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\} + + So \Pr(A) = \frac{18}{36} = \frac{1}{2}, + \Pr(B) = \frac{15}{36} = \frac{5}{12}, and + \Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}. + Finally: + + \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \neq \Pr(A), + + so A and B are not independent. +

+ + + + + +

+ An experiment consists of flipping a fair coin three times. + Let A be the event that the first and second flips match. + Let B be the event that there are at least two heads. + Are A and B independent? +

+ + + +

+ A and B are independent. +

+ + + +

+ + A \amp = \{ HHH, HHT, TTH, TTT \} + B \amp = \{ HHH, HHT, HTH, THH \} + A\cap B \amp = \{HHH, HHT\} + + So \Pr(A) = \frac{4}{8} = \frac{1}{2}, + \Pr(B) = \frac{4}{8} = \frac{1}{2}, and + \Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}. + Finally: + + \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} = \Pr(A), + + so A and B are independent. +

+ + + + + +

+ Let A = \{1, 2, 3\} and B = \{3, 4, 5\} be events in the + sample space \Omega = \{1, 2, 3, 4, 5, 6\}. + Create a probability distribution for \Omega so that + A, B are independent. +

+ + + + + Example Distribution + + + + x + \Pr(x) + + + + 1 + 0.1 + + + + 2 + 0.2 + + + + 3 + 0.2 + + + + 4 + 0.1 + + + + 5 + 0.1 + + + + 6 + 0.3 + + +
+ + + + + Example Distribution + + + + x + \Pr(x) + + + + 1 + 0.1 + + + + 2 + 0.2 + + + + 3 + 0.2 + + + + 4 + 0.1 + + + + 5 + 0.1 + + + + 6 + 0.3 + + +
+ +

+ Now \Pr(A) = 0.5, \Pr(B) = 0.4, and + + \Pr(A\cap B) = 0.2 = (0.5)(0.4) = \Pr(A)\Pr(B), + + so A and B are independent. +

+ + + + + +

+ An experiment consists of flipping a biased coin 20 times. + If the coin comes up heads with probability p = 0.3, find the + probability of seeing 5 heads. + Find the probability of seeing up to (and including) 3 heads.

- Let T be the number of flips until we see heads. - Then T is geometric with parameter p = 0.4, so: + Let S be the number of heads. + Then S \sim \Bin(20, 0.3), so: - \Pr(T = k) \amp = (1-0.4)^{k-1}(0.4) = 0.6^{k-1} \cdot 0.4 + \Pr(S = 5) \amp = {20 \choose 5} (0.3)^5 (0.7)^{20 - 5} + \amp = \frac{20!}{(5!)(15!)} (0.3)^5 (0.7)^{15} + \amp = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} (0.3)^5 (0.7)^{15} + \amp = (19 \times 3 \times 17 \times 16) (0.3)^5 (0.7)^{15} + \amp \approx 0.179

- + + + +

+ An experiment consists of flipping a coin repeatedly until we first + see heads. +

+ - + + +

+ If the coin comes up heads with probability 0.4, what is the + probability we'll see our first heads within three flips? + What about precisely on the third flip? +

+ + + +

+ Let T be the number of flips until we see heads. + Then T is geometric with parameter p = 0.4, so: + + \Pr(T = k) \amp = (1-0.4)^{k-1}(0.4) = 0.6^{k-1} \cdot 0.4 + +

+ + + + + +

+ Which flip has the highest chance of being the first flip to come up + heads? +

+ + + +

+ \Pr(T = k) = 0.6^{k-1} \cdot 0.4, so every additional flip + multiplies the probability by 0.6. + Therefore, the highest value for \Pr(T = k) occurs when + k = 1, in which case \Pr(T = 1) = 0.4. +

+ + + + +

- Which flip has the highest chance of being the first flip to come up heads? + A particular store has an average of 20 customers each hour. + During a 4-hour afternoon shift, what is the probability of serving 80 + customers.

- - - - -

- A particular store has an average of 20 customers each hour. - During a 4-hour afternoon shift, what is the probability of serving 80 customers. -

- - + +

+ Let N be the number of customers seen during the afternoon + shift. + Since the store averages 20 customers per hour, it will average 80 per + 4-hours. + So N \sim \Poiss(20, 4) will have distribution + \Pr(N = k) = \frac{80^{k}}{k!} e^{-80}. + Therefore: + + \Pr(N = 80) = \frac{80^{80}}{80!} e^{-80} \approx 0.045. + +

+ + - - -

- A continuous random variable X taking values in [1, 4] has p.d.f. - f(x) = k(x - \sqrt{x}) for some constant k. -

- + + +

+ A continuous random variable X taking values in [1, 4] + has p.d.f. + f(x) = k(x - \sqrt{x}) for some constant k. +

+ + + +

+ What is the value of k? +

+ - + +

+ k = \frac{6}{17}. +

+ + + + + +

+ Find \Pr(2 \leq X \leq 3). +

+ + + +

+ \Pr(2 \leq X \leq 3) \approx 0.325 +

+ + + + +

- What is the value of k? + A continuous random variable X taking values in [1, 2] + has p.d.f. + \displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}. + Find the c.d.f. + F(x). + Use your c.d.f. + to find \Pr\left(1 \leq X \leq \frac{3}{2}\right).

- k = \frac{6}{17}. + \frac{1}{2}\left(\frac{x^2}{2} - \frac{1}{x}\right) + \frac{1}{4}. + \Pr\left(1 \leq X \leq \frac{3}{2}\right) \approx 0.479.

- + - - +

- Find \Pr(2 \leq X \leq 3). + A continuous random variable X taking values in [2, 3] + has c.d.f. + F(x) = \frac{x^3}{3} - x^2 + 4. + Find the p.d.f. + f(x).

- \Pr(2 \leq X \leq 3) \approx 0.325 + f(x) = x^2 - 2x.

- - + - - -

- A continuous random variable X taking values in [1, 2] has p.d.f. - \displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}. - Find the c.d.f. - F(x). - Use your c.d.f. - to find \Pr\left(1 \leq X \leq \frac{3}{2}\right). -

- - - -

- \frac{1}{2}\left(\frac{x^2}{2} - \frac{1}{x}\right) + \frac{1}{4}. \Pr\left(1 \leq X \leq \frac{3}{2}\right) \approx 0.479. -

- - - - - -

- A continuous random variable X taking values in [2, 3] has c.d.f. - F(x) = \frac{x^3}{3} - x^2 + 4. - Find the p.d.f. - f(x). -

- - - -

- f(x) = x^2 - 2x. -

- - - - - -

- Consider X, Y with the joint distribution table below. - Are X, Y independent? -

- - - Joint distribution for <m>X, Y</m> - - - - - X = 0 - X = 1 - - - - Y = 0 - 0.2 - 0.3 - - - - Y = 1 - 0.4 - 0.1 - - -
- - - -

- No. - For example, \Pr(X = 0, Y = 0) \neq \Pr(X = 0)\Pr(Y = 0). -

- - - - - -

- Suppose X, Y have the distributions: - - \Pr(X = 0) \amp = 0.1 \amp \Pr(Y = 0) \amp = 0.4 - \Pr(X = 1) \amp = 0.4 \amp \Pr(Y = 1) \amp = 0.6 - \Pr(X = 2) \amp = 0.5 - - Assuming X, Y are independent, write a joint distribution table. -

- - - - - Joint Distribution - - - - - X = 0 - X = 1 - X = 2 - - - - Y = 0 - 0.04 - 0.16 - 0.2 - - - - Y = 1 - 0.06 - 0.24 - 0.3 - - -
- - - - - -

- Consider a random variable X with probability distribution below. - Find \E(X). -

- - - - - - - x - \Pr(X = x) - - - - 1 - 0.1 - - - - 2 - 0.05 - - - - 3 - 0.2 - - - - 4 - 0.15 - - - - 5 - 0.15 - - - - 6 - 0.1 - - - - 7 - 0.05 - - - - 8 - 0.1 - - - - 9 - 0.1 - - -
- - - -

- 4.8. -

- - - - - -

- Suppose we flip a coin n = 100 times, and let N count the number of heads. -

- - - - +

- If the coin comes up heads on a flip with probability p = 0.4, what is \E(N)? + Let T \sim \Exp(3). + Find \Pr(1 \leq T \leq 3) and \Pr(T \geq 0.5).

- 40. + \Pr(1 \leq T \leq 3) \approx 0.050. + \Pr(T \geq 0.5) \approx 0.223.

- + - - +

+ Consider X, Y with the joint distribution table below. + Are X, Y independent? +

+ + + Joint distribution for <m>X, Y</m> + + + + + X = 0 + X = 1 + + + + Y = 0 + 0.2 + 0.3 + + + + Y = 1 + 0.4 + 0.1 + + +
+ + + +

+ No. + For example, \Pr(X = 0, Y = 0) \neq \Pr(X = 0)\Pr(Y = 0). +

+ + + + + +

+ Suppose X, Y have the distributions: + + \Pr(X = 0) \amp = 0.1 \amp \Pr(Y = 0) \amp = 0.4 + \Pr(X = 1) \amp = 0.4 \amp \Pr(Y = 1) \amp = 0.6 + \Pr(X = 2) \amp = 0.5 + + Assuming X, Y are independent, write a joint distribution + table. +

+ + + + + Joint Distribution + + + + + X = 0 + X = 1 + X = 2 + + + + Y = 0 + 0.04 + 0.16 + 0.2 + + + + Y = 1 + 0.06 + 0.24 + 0.3 + + +
+ + + + + +

+ Consider a random variable X with probability distribution + below. + Find \E(X). +

+ + + + <tabular halign="center"> + <row header="yes" bottom="minor"> + <cell><m>x</m></cell> + <cell><m>\Pr(X = x)</m></cell> + </row> + + <row> + <cell>1</cell> + <cell>0.1</cell> + </row> + + <row> + <cell>2</cell> + <cell>0.05</cell> + </row> + + <row> + <cell>3</cell> + <cell>0.2</cell> + </row> + + <row> + <cell>4</cell> + <cell>0.15</cell> + </row> + + <row> + <cell>5</cell> + <cell>0.15</cell> + </row> + + <row> + <cell>6</cell> + <cell>0.1</cell> + </row> + + <row> + <cell>7</cell> + <cell>0.05</cell> + </row> + + <row> + <cell>8</cell> + <cell>0.1</cell> + </row> + + <row> + <cell>9</cell> + <cell>0.1</cell> + </row> + </tabular> + </table> + </statement> + + <answer> + <p> + 4.8. + </p> + </answer> + </exercise> + + <exercise> + <statement> + <p> + Let <m>X</m> and <m>Y</m> be random variables each taking the values + 1, 2, 3, 4, 5. + Write different distribution tables for <m>X</m> and <m>Y</m> so that + they have the same expected value. + </p> + </statement> + + <answer> + <table> + <title>Example Distributions + + + + a + \Pr(X = a) + \Pr(Y = a) + + + + 1 + 0.2 + 0.1 + + + + 2 + 0.2 + 0.1 + + + + 3 + 0.2 + 0.6 + + + + 4 + 0.2 + 0.1 + + + + 5 + 0.2 + 0.1 + + +
+ + + + + +

+ Suppose we flip a coin n = 100 times, and let N count + the number of heads. +

+ + + + +

+ If the coin comes up heads on a flip with probability + p = 0.4, what is \E(N)? +

+ + + +

+ 40. +

+ + + + + +

+ What if n = 80 and p = 0.6? +

+ + + +

+ 48. +

+ + + + + +

+ What if n = 200 and p = 0.5? +

+ + + +

+ 100. +

+ + + + + + +

+ If \E(X) = 3, \E(Y) = -2, and \E(Z) = 1, what is + \E(4X + 5Y - Z + 3)? +

+ + + +

+ 4. +

+ + + + + +

+ A continuous random variable X taking values in [0, 1] + has p.d.f. + f(x) = 2x. + What is \E(X)? +

+ + + +

+ 2/3. +

+ + + + + +

+ A continuous random variable X taking values in [-1, 1] + has p.d.f. + f(x) = \frac{3x^2}{2}. + What is \E(X)? +

+ + + +

+ 0. +

+ + + + + +

+ A continuous random variable X taking values in [1, 4] + has p.d.f. + f(x) = \frac{4}{3x^2}. + What is \E(X)? +

+ + + +

+ \frac{4}{3}\ln(4) \approx 1.85. +

+ + + + + +

+ A continuous random variable X taking values in [1, 4] + has p.d.f. + f(x) = k(x - \sqrt{x}) for some constant k. + In a previous problem ( ), + you found the value of k. + Now, find \E(X). +

+ + + +

+ \frac{258}{85} \approx 3.04. +

+ + + + + +

+ A continuous random variable X taking values in [1, 2] + has p.d.f. + \displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}. + Find \E(X). +

+ + + +

+ \frac{1}{2}\left( \ln(2) + \frac{7}{3}\right) \approx 1.513. +

+ + + + + +

+ Consider a random variable X with probability distribution + below. + Find \Var(X). +

+ + + + <tabular halign="center"> + <row header="yes" bottom="minor"> + <cell><m>x</m></cell> + <cell><m>\Pr(X = x)</m></cell> + </row> + + <row> + <cell>1</cell> + <cell>0.1</cell> + </row> + + <row> + <cell>2</cell> + <cell>0.05</cell> + </row> + + <row> + <cell>3</cell> + <cell>0.2</cell> + </row> + + <row> + <cell>4</cell> + <cell>0.15</cell> + </row> + + <row> + <cell>5</cell> + <cell>0.15</cell> + </row> + + <row> + <cell>6</cell> + <cell>0.1</cell> + </row> + + <row> + <cell>7</cell> + <cell>0.05</cell> + </row> + + <row> + <cell>8</cell> + <cell>0.1</cell> + </row> + + <row> + <cell>9</cell> + <cell>0.1</cell> + </row> + </tabular> + </table> + </statement> + + <answer> + <p> + <m>5.76</m>. + </p> + </answer> + </exercise> + + <exercise> + <statement> + <p> + Let <m>X</m> be a random variable taking the values 1, 2, 3, 4, 5. + Write a distribution table for <m>X</m>, then use your table to write + a distribution for <m>X^2</m>. + Then, find <m>\Var(X)</m>. + </p> + </statement> + </exercise> + + <exercise> + <statement> + <p> + Suppose we flip a coin <m>n = 100</m> times, and let <m>N</m> count + the number of heads. + If the coin comes up heads on a flip with probability <m>p = 0.3</m>, + what is <m>\Var(N)</m>? What if <m>n = 80</m> and <m>p = 0.6</m>? - </p> - </statement> - - <answer> - <p> - 48. - </p> - </answer> - </task> - - - <task> - <statement> - <p> What if <m>n = 200</m> and <m>p = 0.5</m>? </p> </statement> <answer> <p> - 100. + When <m>n = 100</m> and <m>p = 0.3</m>, <m>\Var(N) = 21</m>. + When <m>n = 80</m> and <m>p = 0.6</m>, <m>\Var(N) = 19.2</m>. + When <m>n = 200</m> and <m>p = 0.5</m>, <m>\Var(N) = 50</m>. </p> </answer> - </task> - </exercise> + </exercise> - <exercise> - <statement> - <p> - If <m>\E(X) = 3</m>, <m>\E(Y) = -2</m>, and <m>\E(Z) = 1</m>, what is <m>\E(4X + 5Y - Z + 3)</m>? - </p> - </statement> + <exercise> + <statement> + <p> + If <m>\E(X) = 3</m>, <m>\Var(X) = 2</m>, what is <m>\E(X^2)</m>? + </p> + </statement> - <answer> - <p> - 4. - </p> - </answer> - </exercise> + <answer> + <p> + <m>11</m>. + </p> + </answer> + </exercise> - <exercise> - <statement> - <p> - A continuous random variable <m>X</m> taking values in <m>[0, 1]</m> has p.d.f. - <m>f(x) = 2x</m>. - What is <m>\E(X)</m>? - </p> - </statement> + <exercise> + <statement> + <p> + A continuous random variable <m>X</m> taking values in <m>[0, 1]</m> + has p.d.f. + <m>f(x) = 2x</m>. + What is <m>\Var(X)</m>? + </p> + </statement> - <answer> - <p> - 2/3. - </p> - </answer> - </exercise> + <answer> + <p> + <m>1/18</m>. + </p> + </answer> + </exercise> - <exercise> - <statement> - <p> - A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> has p.d.f. - <m>f(x) = \frac{4}{3x^2}</m>. - What is <m>\E(X)</m>? - </p> - </statement> + <exercise> + <statement> + <p> + A continuous random variable <m>X</m> taking values in <m>[-1, 1]</m> + has p.d.f. + <m>f(x) = \frac{3x^2}{2}</m>. + What is <m>\Var(X)</m>? + </p> + </statement> - <answer> - <p> - <m>\frac{4}{3}\ln(4) \approx 1.85.</m> - </p> - </answer> - </exercise> + <answer> + <p> + <m>0.6</m>. + </p> + </answer> + </exercise> - <exercise> - <statement> - <p> - A continuous random variable <m>X</m> taking values in <m>[1, 2]</m> has p.d.f. - <m>\displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}</m>. - Find <m>\E(X)</m>. - </p> - </statement> + <exercise> + <statement> + <p> + A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> + has p.d.f. + <m>f(x) = \frac{4}{3x^2}</m>. + What is <m>\Var(X)</m>? + </p> + </statement> - <answer> - <p> - <m>\frac{1}{2}\left( \ln(2) + \frac{7}{3}\right) \approx 1.513.</m> - </p> - </answer> - </exercise> + <answer> + <p> + <m>4 - \left(\frac{4}{3}\ln(4)\right)^2 \approx 0.583</m>. + </p> + </answer> + </exercise> - <exercise> - <statement> - <p> - Consider a random variable <m>X</m> with probability distribution below. - Find <m>\Var(X)</m>. - </p> + <exercise> + <statement> + <p> + A continuous random variable <m>X</m> taking values in <m>[1, 4]</m> + has p.d.f. + <m>f(x) = k(x - \sqrt{x})</m> for some constant <m>k</m>. + In a previous problem ( <xref ref="exercise-continuous-RV-find-k"/> ), + you found the value of <m>k</m>. + Now, find <m>\Var(X)</m>. + </p> + </statement> - <table> - <title> + +

+ \frac{2307}{238} - \left(\frac{258}{85}\right)^2 \approx 0.48. +

+ + - - - x - \Pr(X = x) - + + +

+ A continuous random variable X taking values in [1, 2] + has p.d.f. + \displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}. + Find \Var(X). +

+ - - 1 - 0.1 - + +

+ \frac{29}{24} - \frac{1}{2}\ln(2) \approx 0.862. +

+ + - - 2 - 0.05 - + + +

+ Suppose a coin has an unknown probability of coming up heads. + We perform the experiment in five independent trials, during which it + takes 4, 5, 4, 3, and 6 flips to see our first heads in each trial. + What is the maximum likelihood estimation for the probability of the + coin coming up heads on a flip? +

+ + - - 3 - 0.2 - + - - 5 - 0.15 - + + +

+ A particular store owner wants to approximate the average hourly rate + at which customers come into the store. + They observe 80 customers enter during a particular 4-hour shift. + What is the maximum likelihood estimation for the hourly customer + rate? +

+ + - - 6 - 0.1 - + + +

+ A radioactive material emits particles at an unknown probabilistic + rate \lambda particles per minute. + We observe particles emitted at times 1.1, 1.7, 1.3, 2.2, 1.9, and 1.8 + minutes. + Write the likelihood function \mathcal{L}(\lambda) based on + this data. + What is the maximum likelihood estimation for \lambda? +

+ + - - 7 - 0.05 - - - - 8 - 0.1 - - - - 9 - 0.1 - - -
- - - - - -

- Suppose we flip a coin n = 100 times, and let N count the number of heads. - If the coin comes up heads on a flip with probability p = 0.4, what is \Var(N)? What if n = 80 and p = 0.6? What if n = 200 and p = 0.5? -

- - - - - -

- If \E(X) = 3, \Var(X) = 2, what is \E(X^2)? -

- - - - - -

- A continuous random variable X taking values in [0, 1] has p.d.f. - f(x) = 2x. - What is \Var(X)? -

- - - - - -

- A continuous random variable X taking values in [1, 4] has p.d.f. - f(x) = \frac{4}{3x^2}. - What is \Var(X)? -

- - - - - -

- A continuous random variable X taking values in [1, 2] has p.d.f. - \displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}. - Find \Var(X). -

- - - - - -

- Suppose a coin has an unknown probability of coming up heads. - We perform the experiment in n independent trials, during which it takes k_1, k_2, \dotsc, k_n flips to see our first heads in each trial. - Find a "common sense" MLE formula for the geometric distribution. -

- - - - - -

- A particular store owner wants to approximate the average hourly rate at which customers come into the store. - They observe 80 customers enter during a particular 4-hour shift. - What is the maximum likelihood estimation for the hourly customer rate? -

- - - - - -

- A radioactive material emits particles at an unknown probabilistic rate \lambda particles per minute. - We observe particles emitted at times 1.1, 1.7, 1.3, 2.2, 1.9, and 1.8 minutes. - Write the likelihood function \mathcal{L}(\lambda) based on this data. - What is the maximum likelihood estimation for \lambda? -

- - - - - -

- Suppose a parameter \theta takes values in [0, 1] with likelihood function \mathcal{L}(\theta) = \sqrt{\theta} - \theta^2. - Find the maximum likelihood estimation of \theta. -

- - + + +

+ Suppose a parameter \theta takes values in [0, 1] with + likelihood function + \mathcal{L}(\theta) = \sqrt{\theta} - \theta^2. + Find the maximum likelihood estimation of \theta. +

+ + \ No newline at end of file