From 0e69c12d7e01560b25124addc9073b73c8883c15 Mon Sep 17 00:00:00 2001 From: andyeisenberg Date: Tue, 10 Feb 2026 10:55:39 -0500 Subject: [PATCH] Exam 1 Review --- source/review/Exam-1-Review.ptx | 1094 +++++++++++++++++++++++++++++++ 1 file changed, 1094 insertions(+) diff --git a/source/review/Exam-1-Review.ptx b/source/review/Exam-1-Review.ptx index aed9e5f..8bc9e26 100644 --- a/source/review/Exam-1-Review.ptx +++ b/source/review/Exam-1-Review.ptx @@ -29,8 +29,1102 @@ + +

+ Consider the sets A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, B = \{2, 4, 9, 10, 12, 14, 19\}, and C = \{9, 10, 11, 14, 16, 17, 20\}, which are all subsets of \Omega = \{1, 2, 3, \dotsc, 20\}. +

+ + + + + +

+ Find A - (B \cap C). +

+ + + +

+ \{1, 2, 3, 4, 5, 6, 7, 8\} +

+ + + + + + +

+ Find |A|, |B|, |C|, |A\cup B|, |A \cap B|, |B\cap C|, |A\cap C|, and |A\cup B\cup C|. + Is it true that the size of the union of sets is equal to the sum of the sizes of the individual sets? +

+ + + +

+ |A| = 10, |B| = 7, |C| = 7, |A \cup B| = 13, |A \cap B| = 4, |B\cap C| = 3, |A\cap C| = 2, |A\cup B\cup C| = 17. In particular, note that |A\cup B| = 13 \neq 10 + 7 = |A| + |B|, so it is not true in general that the size of the union of sets is the sum of the sizes of the individual sets. +

+ + + + + + +

+ Find A^c and (A\cup B)^c. +

+ + + +

+ A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}, (A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}. +

+ + + + + + +

+ 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.] +

+ + + +

+ + \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. + So the answer here is the same as in . +

+ + + + + +

+ 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 = \{ HHH, HHT, HTH, THH, 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? +

+ + + +

+ 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. +

+ + + + + + + x + \Pr(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 + + +
+ + + +

+ \Pr(A) = 0.45, \Pr(B) = 0.4, \Pr(A \cup B) = 0.6, \Pr(A \cap B) = 0.25. +

+ + + + + +

+ 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 + + +
+ + + + + +

+ 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 + + +
+ + + + + +

+ 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 nth minute. + What is the probability of the molecule leaving the cell during the first 3 minutes? +

+ + + +

+ For short, write \Pr(n) to mean the probability of the toxin molecule leaving during the nth minute. + Then \Pr(n) = (0.7)^{n - 1} (0.3). +

+ +

+ The probability of leaving during the first 3 minutes is \Pr(1) + \Pr(2) + \Pr(3) = 0.657. +

+ + + + + + +

+ 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? +

+ + + +

+ + \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 = \{ \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? +

+ + + +

+ 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? +

+ + + + + + +

+ 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 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 +

+ + + + + + +

+ 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)? +

+ + + +

+ 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, 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, 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). +

+ + + + + + + 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 + + +
+ + + + + +

+ 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). +

+ \ No newline at end of file