Midterm Info + Prep

Midterm Information

  • The Midterm is scheduled for in class on Tuesday June 2
  • It is 1 hour and 50 minutes
  • Your midterm will cover materials from Lectures 1 - 8
  • You may bring in one (1) “cheat sheet”:
    • Must be HAND WRITTEN with pen/pencil on said sheet of paper (not typed, not photo copied, not printed, not written on an iPad)
    • Must be on 8.5 by 11 inch sheet of paper or smaller d
    • You may write on both sides
    • No magnifying glasses or anything else silly
    • I will confiscate cheatsheets that do not follow these rules 🥀
    • I do not care what is written on it
  • Exam is hand written on paper, bring something to write with
  • You may bring a non-programmable, non-graphing calculator.

Midterm Preparation

Here are some selected problems that would serve as good practice for the midterm.

Question 1 (Conditional Probability/Combinatorics)

There are two boxes. The first box has 4 blue balls and 5 green balls, and the second box has 3 blue balls and 2 green balls. One ball is randomly drawn from the first box and put into the second box. Then a ball is randomly drawn from the second box. What is the conditional probability that the transferred ball from the first box to the second box is blue given that a blue ball is drawn from the second box?

Let \(B_2\) be the event “a blue ball is drawn from the second box”, and \(B_1\) the event “a blue ball is drawn from the first box and put in the second”. We want \(\mathbb{P}(B_1 \ \vert\ B_2)\).

We have \(\mathbb{P}(B_1 \ \vert\ B_2) = \frac{\mathbb{P}(B_2\ \vert\ B_1)\mathbb{P}(B_1)}{\mathbb{P}(B_2)}\)

Also, \(\mathbb{P}(B_1) = 4/9\) and \(\mathbb{P}(B_2 \ \vert\ B_1) = 4/6\). Finally, if \(G_1\) is the event “a green ball is drawn from the first box and put in the second” (by the way, note that \(G_1 = B_1^c\)), we have:

\[ \begin{aligned} \mathbb{P}(B_2) &= \mathbb{P}(B_2 \ \vert\ B_1) \mathbb{P}(B_1) + \mathbb{P}(B_2 \ \vert\ B_1^c)\mathbb{P}(B_1^c)\\ &=\mathbb{P}(B_2 \ \vert\ B_1) \mathbb{P}(B_1) + \mathbb{P}(B_2 \ \vert\ G_1)\mathbb{P}(G_1)\\ &= (4/6) (4/9) + (3/6) (5/9)\\ &= 31/54 \end{aligned} \]

Putting it all together, we get: \[ \begin{aligned} \mathbb{P}(B_1 \ \vert\ B_2) &= \frac{(4/6)(4/9)}{31/54} = 16/31. \end{aligned} \]

Question 2 (Set Theory)

Let \(A\), \(B\) and \(C\) be events (sets) in a sample space \(\Omega\).

  1. Prove that if \(A\), \(B\) and \(C\) are independent then \(A \cup B\) and \(C\) are also independent.

We will prove it using the definition. We need to verify that \(P((A \cup B) \cap C) = P(A \cup B)P(C)\).

Note that \((A \cup B) \cap C = (A \cap C) \cup (B \cap C)\), hence:

\[ \begin{aligned} P((A \cup B) \cap C) &= P((A \cap C) \cup (B \cap C)) \\ &= P(A \cap C) + P(B \cap C) - P((A \cap C) \cap (B \cap C)) \\ &= P(A \cap C) + P(B \cap C) - P(A \cap B \cap C) \\ &= P(A)P(C) + P(B)P(C) - P(A)P(B)P(C) \\ &= P(C) \big( P(A) + P(B) - P(A)P(B) \big) \\ &= P(C) \big( P(A) + P(B) - P(A \cap B) \big) \\ &= P(C)P(A \cup B) \end{aligned} \]

  1. Prove that if \(\mathbb{P}(A) > 0\) and \(\mathbb{P}(A \cap B) > 0\) then \(\mathbb{P}(A \cap B \cap C) = P (C \ \vert\ A \cap B) \mathbb{P}(B\ \vert\ A) \mathbb{P}(A)\).

The right hand side is:

\[ \begin{aligned} P(C \mid A \cap B) P(B \mid A) P(A) &= \frac{P(C \cap (A \cap B))}{P(A \cap B)} \cdot \frac{P(B \cap A)}{P(A)} \cdot P(A) \\ &= P(C \cap A \cap B) \\ &= P(A \cap B \cap C) \end{aligned} \]

Question 3 (Combinatorics)

Consider a standard set of poker cards (52 cards, 13 of each of four suits (diamonds, hearts, spades and clubs)). The cards of each suit are numbered 2 to 10, J, Q, K and A. A hand consists of 5 randomly chosen cards. A “full house” consists of a hand with two cards with the same face value, and the other three cards also sharing a face value. For example, (8, 2, 8, 2, 2) is a “full house”, but (Q, 9, Q, Q, Q) is not. If all hands are equally likely, what is the probability of receiving a “full house”?

The sample space consists of all possible hands of 5 randomly chosen cards. There are \(\binom{52}{5}\) of them. Those that are full houses can be listed as follows: first choose which face values will be in the full house, there are 13 choices for the face value for the pair, and then \(\binom{4}{2}\) ways of selecting 2 of the 4 cards with that value. For each of those combinations we have 12 choices for the face of the triplet, and \(\binom{4}{3}\) ways of selecting the 3 cards of that value.

Thus, the probability is:

\[ \frac{13 \times \binom{4}{2} \times 12 \times \binom{4}{3}}{\binom{52}{5}} = 0.001440576 \]

Question 4 (Permutations)

How many different arrangements of the letters in the word “singing” can be obtained?

There are \(7\) letters, so there are \(7!\) possible arrangements if we distinguish the i’s, g’s and n’s as different. However, any permutation of these pairs of letters will not change the overall letter arrangement. So we are overcounting each pattern by a factor of \(2! × 2! × 2! = 8\). Thus, the number of different letter arrangements is \(7! / 8 = 630.\)

Question 5 (Discrete Random Variables)

An urn contains 4 red balls and 2 blue balls. A student randomly selects balls one at a time, without replacement, until they select a blue ball. Let the random variable \(X\) denote the total number of balls selected. Calculate the probability mass function (PMF), \(P(X = x)\), for all possible values of \(x\). Express your final answer as a single equation using indicators.

Support: \(x \in \{1, 2, 3, 4, 5\}\)

For \(X = 1\): The first ball is blue. \[P(X = 1) = \frac{2}{6} = \frac{1}{3}\]

For \(X = 2\): The first ball is red, and the second is blue. \[ P(X = 2) = \frac{4}{6} \times \frac{2}{5} = \frac{8}{30} = \frac{4}{15} \]

For \(X = 3\): The first two balls are red, and the third is blue.

\[P(X = 3) = \frac{4}{6} \times \frac{3}{5} \times \frac{2}{4} = \frac{24}{120} = \frac{1}{5}\]

For \(X = 4\): The first three balls are red, and the fourth is blue. \[P(X = 4) = \frac{4}{6} \times \frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} = \frac{48}{360} = \frac{2}{15}\]

For \(X = 5\): The first four balls are red, and the fifth is blue. \[P(X = 5) = \frac{4}{6} \times \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} = \frac{48}{720} = \frac{1}{15}\] Therefore:

\[P(X = x) = \frac{5}{15}I_{\{1\}}(x) + \frac{4}{15}I_{\{2\}}(x) + \frac{3}{15}I_{\{3\}}(x) + \frac{2}{15}I_{\{4\}}(x) + \frac{1}{15}I_{\{5\}}(x)\]

Question 6 (Discrete Random Variables)

A researcher is studying the nesting habits of a specific population of waterfowl. Based on historical data, the probability that a randomly selected nest contains at least one unhatched egg at the end of the breeding season is \(0.20\). The researcher randomly samples \(15\) distinct nests from this population. Assume each sample is independent.

  1. What is probability that exactly \(3\) of the sampled nests contain at least one unhatched egg?

  2. What is the probability that less than \(2\) of the sampled nests contain at least one unhatched egg.

\(Y \sim \text{Binomial}(n = 15, p = 0.20)\). Therefore the PMF is: \[P(Y = y) = \binom{15}{y} (0.20)^y (0.80)^{15-y} \quad \text{for } y \in \{0, 1, 2, \dots, 15\}\]

\[P(Y = 3) = \binom{15}{3} (0.20)^3 (0.80)^{12} = 0.2501\]

\[P(Y < 2) = P(Y = 0) + P(Y = 1)\] Then \[P(Y = 0) = \binom{15}{0} (0.20)^0 (0.80)^{15} = 1 \times 1 \times 0.03518 \approx 0.0352\] and \[P(Y = 1) = \binom{15}{1} (0.20)^1 (0.80)^{14} = 15 \times 0.20 \times 0.04398 \approx 0.1319\]

Therefore: \[P(Y < 2) = 0.0352 + 0.1319 = 0.1671\]

Question 7 (Discrete Random Variables)

A biology department receives a shipment of \(20\) field-grade digital thermometers. Unbeknownst to the lab manager, \(5\) of these thermometers are improperly calibrated. A teaching assistant randomly selects \(6\) thermometers from the shipment (without replacement) to distribute to an upcoming ecology lab section.

  1. Let \(X\) be the number of improperly calibrated thermometers from the sample of 6. What distribution does \(X\) follow? Name it, and identify the relevant parameters.

\(X\) follows a hypergeometric distribution with:

  • Population Size: \(N = 20\)
  • Number of Successes (total improperly calibrated thermometers): \(K = 5\)
  • Sample Size \(n = 6\)

The probability mass function (PMF) is given by:\[P(X = x) = \frac{\binom{K}{x} \binom{N - K}{n - x}}{\binom{N}{n}} = \frac{\binom{5}{x} \binom{15}{6 - x}}{\binom{20}{6}}\]

  1. Calculate the probability that at most \(1\) of the selected thermometers is improperly calibrated.

\[P(X \le 1) = P(X = 0) + P(X = 1)\].

\[P(X = 0) = \frac{\binom{5}{0} \binom{15}{6}}{\binom{20}{6}} = \frac{1 \times 5005}{38760} \approx 0.1291\].

\[P(X = 1) = \frac{\binom{5}{1} \binom{15}{5}}{\binom{20}{6}} = \frac{5 \times 3003}{38760} = \frac{15015}{38760} \approx 0.3874\]

Therefore, \[P(X \le 1) = 0.1291 + 0.3874 = 0.5165\]

Question 8 (Continuous Random Variables)

Consider the probability density function (PDF) of \(X\) given by: \[f_X(x) = \begin{cases} c(4x - x^3) & \text{for } 0 \le x \le 2 \\ 0 & \text{otherwise} \end{cases}\].

  1. What value of \(c\) makes \(f_X(x)\) a valid probability density function (PDF)?

In order for the PDF to be valid, we must have:

\[\int_{-\infty}^{\infty} f(x) \, dx = 1 \implies \int_{0}^{2} c(4x - x^3) \, dx = 1\]

which means \[c \left[ 2x^2 - \frac{x^4}{4} \right]_{0}^{2} = 1\] \[c \left[ \left( 2(2)^2 - \frac{(2)^4}{4} \right) - (0) \right] = 1\] \[c \left[ 8 - \frac{16}{4} \right] = 1\]

This means \[c [8 - 4] = 1 \implies 4c = 1 \implies c = \frac{1}{4}\]

  1. Calculate \(\mathbb{P}(X > 1)\)

\[P(X > 1) = \int_{1}^{2} \frac{1}{4}(4x - x^3) \, dx = \frac{1}{4} \left[ 2x^2 - \frac{x^4}{4} \right]_{1}^{2}\]

\[= \frac{1}{4} \left[ \left( 2(2)^2 - \frac{(2)^4}{4} \right) - \left( 2(1)^2 - \frac{(1)^4}{4} \right) \right] = 9/16\]

Question 9 (Continuous Random Variables)

A marine biology station monitors the breaches of humpback whales within a designated conservation bay. Based on long-term survey data, whales breach in this bay at a constant average rate of \(2.4\) times per hour.

Assuming whale breaches occur independently and at a constant rate over time. Let the continuous random variable \(T\) represent the time (in hours) between consecutive whale breaches. C

Calculate the probability that the waiting time between two consecutive whale breaches is less than \(15\) minutes.

\(T \sim {\mathrm{Exp}}(\lambda = 2.4)\). The PDF is \[f_T(t) = \begin{cases} 2.4e^{-2.4t} & \text{for } t \ge 0 \\ 0 & \text{otherwise} \end{cases}\]

NOTE: 15 minutes = 0.25 hours

\[ \begin{aligned} \mathbb{P}(T\ le 0.25) = \int_{0}^{0.25}2.4e^{-2.4t}dt = 1 - e^{-2.4(0.25)} = 0.4512 \end{aligned} \]

Question 10 (Transformations)

Coming soon

Question 11 (Transformations)

Coming soon

Question 12 (Joint Distributions)

Coming soon

Question 13 (Marginal Distributions)

Coming soon