Exam 1 Review

Use the following problems to prepare for the exam. There will be in-class review on Thursday, February 12. Your recitation this week will also be exam review.

Allowed Materials

You will be allowed to use a scientific calculator (not a graphing calculator, not a calculator app on your phone). You may not share a calculator with another student; you must use your own calculator.

You may bring a standard 3 in x 5 in index card with prepared notes. You may use both sides of the notecard. You must put your full name in the top right corner of the card, and turn it in along with your exam.

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?

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?

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)?

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).

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.