IIT Madras
Statistics II
Graded Assignment 1

Graded Assignment 1

alt text alt text alt text alt text alt text

alt text alt text alt text

alt text alt text

Of course! Here are the solutions and explanations for all the questions you provided.

Question 1: Conditional Probability

Problem: The joint PMF of two discrete random variables $X$ and $Y$ is given by: $f_{XY}(x, y) = \begin{cases} \frac{1}{36}(3x + y), & x, y \in {0, 1, 2} \ 0, & \text{otherwise} \end{cases}$ What is the value of $P(0 < X < 2 \ | \ Y > 1)$? Enter the answer correct to two decimal places.

Answer: 0.33

Concepts Explained 💡

  • Joint Probability Mass Function (PMF): A function $f_{XY}(x, y)$ that gives the probability that discrete random variables $X$ and $Y$ take on the specific values $x$ and $y$, respectively. So, $f_{XY}(x, y) = P(X=x, Y=y)$.
  • Marginal PMF: The probability distribution of a single variable from a joint distribution. For example, to get the marginal PMF of $Y$, $f_Y(y)$, you sum the joint probabilities over all possible values of $X$: $f_Y(y) = P(Y=y) = \sum_x f_{XY}(x, y)$.
  • Conditional Probability: The probability of an event occurring, given that another event has already occurred. The formula is $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

Step-by-Step Solution

  1. Interpret the Probability Statement:

    • The condition $0 < X < 2$ for an integer $X$ means $X=1$.
    • The condition $Y > 1$ for an integer $Y$ in the set ${0, 1, 2}$ means $Y=2$.
    • So, we need to calculate $P(X=1 \ | \ Y=2)$.

  2. Apply the Conditional Probability Formula:

    $$P(X=1 \ | \ Y=2) = \frac{P(X=1, Y=2)}{P(Y=2)}$$

  3. Calculate the Joint Probability $P(X=1, Y=2)$: This is given directly by the joint PMF $f_{XY}(1, 2)$.

    $$P(X=1, Y=2) = f_{XY}(1, 2) = \frac{1}{36}(3(1) + 2) = \frac{5}{36}$$

  4. Calculate the Marginal Probability $P(Y=2)$: We need to sum the joint probabilities over all possible values of $X$ (which are 0, 1, and 2) while keeping $Y=2$.

    $$P(Y=2) = \sum_{x=0}^{2} f_{XY}(x, 2) = f_{XY}(0, 2) + f_{XY}(1, 2) + f_{XY}(2, 2)$$

    $$P(Y=2) = \frac{1}{36}(3(0) + 2) + \frac{1}{36}(3(1) + 2) + \frac{1}{36}(3(2) + 2)$$

    $$P(Y=2) = \frac{2}{36} + \frac{5}{36} + \frac{8}{36} = \frac{15}{36}$$

  5. Calculate the Final Conditional Probability:

    $$P(X=1 \ | \ Y=2) = \frac{5/36}{15/36} = \frac{5}{15} = \frac{1}{3}$$

    Converting to a decimal rounded to two places, we get 0.33.

Question 2: Conditional PMF from a Table

Problem: The joint distribution of two random variables $X$ and $Y$ is given as:

Y \ X01
11/41/8
21/4k
301/8

Find the value of $f_{Y|X=1}(2)$.

Answer: 0.50

Concepts Explained 💡

  • Joint Distribution Table: A table that displays the probabilities for each possible pair of outcomes for two random variables.
  • Normalization Property: The sum of all probabilities in a joint distribution must equal 1.
  • Conditional PMF: The probability distribution of one variable given a specific value of another. It’s calculated as: $$f_{Y|X}(y|x) = P(Y=y \ | \ X=x) = \frac{P(X=x, Y=y)}{P(X=x)}$$

Step-by-Step Solution

  1. Find the value of $k$: The sum of all probabilities in the table must be 1.

    $$\frac{1}{4} + \frac{1}{4} + 0 + \frac{1}{8} + k + \frac{1}{8} = 1$$

    $$\frac{2}{4} + \frac{2}{8} + k = 1$$

    $$\frac{1}{2} + \frac{1}{4} + k = 1$$

    $$\frac{3}{4} + k = 1 \implies k = \frac{1}{4}$$

  2. Identify the Required Probability: We need to find $f_{Y|X=1}(2)$, which is the conditional probability $P(Y=2 \ | \ X=1)$.

  3. Apply the Formula:

    $$P(Y=2 \ | \ X=1) = \frac{P(X=1, Y=2)}{P(X=1)}$$

  4. Find the Probabilities from the Table:

    • The joint probability $P(X=1, Y=2)$ is the value of $k$, which we found to be 1/4.
    • The marginal probability $P(X=1)$ is the sum of all probabilities in the column where $X=1$. $$P(X=1) = P(X=1, Y=1) + P(X=1, Y=2) + P(X=1, Y=3)$$ $$P(X=1) = \frac{1}{8} + \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{4} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

  5. Calculate the Result:

    $$P(Y=2 \ | \ X=1) = \frac{1/4}{1/2} = \frac{1}{4} \times 2 = \frac{1}{2}$$

    The final answer is 0.50.

Question 3: Law of Total Probability

Problem: Akshat draws a card from a well-shuffled pack of 52 cards. If the drawn card is a face card, he draws two balls randomly from bag A which contains 5 Red, 6 Black and 4 Green balls. If the drawn card is not a face card, then he draws three balls randomly from bag B which contains 7 Red, 9 Black and 4 Green balls. Let two random variables $X$ and $Y$ be defined as: $X = \begin{cases} 1 & \text{if the drawn card is a face card} \ 0 & \text{if the drawn card is not a face card} \end{cases}$ and $Y$ be the number of Red balls drawn. Find the value of $f_Y(1)$. Write your answer correct up to two decimal places.

Answer: 0.48

Concepts Explained 💡

  • Law of Total Probability: This law allows you to find the probability of an event by considering all possible scenarios (partitions) that lead to it. The formula is: $$P(A) = \sum_i P(A|B_i)P(B_i)$$
  • Combinations (Hypergeometric Probability): Used to find the probability of drawing a specific number of successes (e.g., red balls) without replacement. The formula to choose $k$ successes from a population of $K$ successes in $n$ draws from a total population $N$ is: $$P(\text{k successes}) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$$

Step-by-Step Solution

  1. Identify the Goal: We need to find $f_Y(1)$, which is the overall probability $P(Y=1)$.

  2. Use the Law of Total Probability: The event $Y=1$ (drawing 1 red ball) can happen in two ways: either a face card was drawn first ($X=1$) or not ($X=0$).

    $$P(Y=1) = P(Y=1 | X=1)P(X=1) + P(Y=1 | X=0)P(X=0)$$

  3. Calculate Probabilities for $X$: A standard deck has 12 face cards (J, Q, K).

    • $P(X=1) = P(\text{face card}) = \frac{12}{52} = \frac{3}{13}$
    • $P(X=0) = P(\text{not face card}) = 1 - \frac{3}{13} = \frac{10}{13}$

  4. Calculate Conditional Probabilities for $Y=1$:

    • Case 1: Given $X=1$ (drew a face card, use Bag A) Bag A has 5 Red and 10 non-Red balls (total 15). We draw 2 balls. $$P(Y=1|X=1) = \frac{\text{Ways to choose 1R and 1 non-R}}{\text{Total ways to choose 2 balls}} = \frac{\binom{5}{1}\binom{10}{1}}{\binom{15}{2}}$$ $$= \frac{5 \times 10}{105} = \frac{50}{105} = \frac{10}{21}$$
    • Case 2: Given $X=0$ (drew non-face card, use Bag B) Bag B has 7 Red and 13 non-Red balls (total 20). We draw 3 balls. $$P(Y=1|X=0) = \frac{\text{Ways to choose 1R and 2 non-R}}{\text{Total ways to choose 3 balls}} = \frac{\binom{7}{1}\binom{13}{2}}{\binom{20}{3}}$$ $$= \frac{7 \times 78}{1140} = \frac{546}{1140} = \frac{91}{190}$$

  5. Combine Everything:

    $$P(Y=1) = \left(\frac{10}{21} \times \frac{3}{13}\right) + \left(\frac{91}{190} \times \frac{10}{13}\right)$$

    $$P(Y=1) = \frac{10}{91} + \frac{910}{2470} = \frac{10}{91} + \frac{91}{247} = \frac{10}{91} + \frac{7 \times 13}{19 \times 13} = \frac{10}{91} + \frac{7}{19}$$

    $$P(Y=1) = \frac{10(19) + 7(91)}{91 \times 19} = \frac{190 + 637}{1729} = \frac{827}{1729} \approx 0.4783$$

    Rounding to two decimal places, the answer is 0.48.

Question 4: Fundamental Probability Formulas

Problem: Which of the following options is/are always correct?

  • $f_{XYZ}(x, y, z) = f_X(Y=y, Z=z|x)f_X(x)$
  • $f_X(x) = \sum_{y \in R_Y} f_{XY}(x,y)$
  • $f_{XYZ}(x, y, z) = f_X(Y=y, Z=z|x)f_Y(y)f_Z(z)$
  • $f_{XYZ}(x, y, z) = f_{X|Y=y,Z=z}(x)f_Y(y)f_Z(z)$

Answer:

The only statement that is always correct is:

$$f_X(x) = \sum_{y \in R_Y} f_{XY}(x,y)$$

Concepts Explained 💡

  • Chain Rule of Probability: This rule allows you to break down a joint probability into a product of conditional and marginal probabilities. For three variables, it is: $f_{XYZ}(x,y,z) = f_{Z|X,Y}(z|x,y) f_{Y|X}(y|x) f_X(x)$. Another common form is $f_{XYZ}(x,y,z) = f_{X|Y,Z}(x|y,z) f_{YZ}(y,z)$.
  • Marginal Distribution: As explained before, this is the probability distribution of a subset of random variables from a larger set. The provided formula is the definition of the marginal PMF of $X$ derived from the joint PMF of $X$ and $Y$. It is obtained by “summing out” the other variable, $Y$.
  • Statistical Independence: Two random variables $Y$ and $Z$ are independent if and only if their joint PMF is the product of their marginal PMFs: $f_{YZ}(y,z) = f_Y(y)f_Z(z)$.

Analysis of Options

  1. $f_{XYZ}(x, y, z) = f_X(Y=y, Z=z|x)f_X(x)$: The notation is non-standard, but it appears to be an attempt at the chain rule. A correct version of the chain rule is $f_{XYZ}(x, y, z) = f_{Y,Z|X}(y, z|x)f_X(x)$. While this rule is always true, the notation in the option is garbled and incorrect.
  2. $f_X(x) = \sum_{y \in R_Y} f_{XY}(x,y)$: This is the definition of the marginal PMF of $X$. It is always true for any pair of discrete random variables $X$ and $Y$.
  3. $f_{XYZ}(x, y, z) = f_X(Y=y, Z=z|x)f_Y(y)f_Z(z)$: This is incorrect. It mixes a conditional probability with marginals in a way that is not generally valid.
  4. $f_{XYZ}(x, y, z) = f_{X|Y=y,Z=z}(x)f_Y(y)f_Z(z)$: This is also incorrect. The correct chain rule would be $f_{XYZ}(x,y,z) = f_{X|Y,Z}(x|y,z) f_{Y,Z}(y,z)$. The formula in the option is only true if $Y$ and $Z$ are independent, which is not always the case.

Therefore, the only universally correct statement is the second one.

Question 5: Solving for Parameters in a Joint PMF

Problem: Two random variables $X$ and $Y$ are jointly distributed with joint pmf $f_{XY}(x, y) = a(bx + y)$, where $x$ and $y$ are integers in $0 \le x \le 2$ and $0 \le y \le 3$ such that $P(X > 1, Y < 2) = \frac{4}{9}$. Find the value of $f_{XY}(2, 0)$. Write your answer correct to two decimal places.

Answer: 0.23

Concepts Explained 💡

  • Properties of a Joint PMF:
    1. $f_{XY}(x, y) \ge 0$ for all $(x, y)$.
    2. The sum of the PMF over all possible pairs $(x, y)$ must be 1 (Normalization). $\sum_x \sum_y f_{XY}(x,y) = 1$.
  • Solving Systems of Equations: This problem requires setting up two equations based on the properties of the PMF and the given information, then solving for the unknown parameters a and b.

Note: This problem statement is technically flawed as it leads to a PMF with some negative probabilities. However, we can solve it formally by following the mathematical steps.

Step-by-Step Solution

  1. Set up the Normalization Equation: The sum of $f_{XY}(x, y)$ over all $x \in {0, 1, 2}$ and $y \in {0, 1, 2, 3}$ must be 1.

    $$\sum_{x=0}^{2} \sum_{y=0}^{3} a(bx + y) = 1$$

    $$a \sum_{x=0}^{2} \left( \sum_{y=0}^{3} bx + \sum_{y=0}^{3} y \right) = 1$$

    The inner sum is over 4 values of y, and $\sum_{y=0}^{3} y = 0+1+2+3=6$.

    $$a \sum_{x=0}^{2} (4bx + 6) = 1$$

    $$a [(4b(0)+6) + (4b(1)+6) + (4b(2)+6)] = 1$$

    $$a [6 + 4b+6 + 8b+6] = 1$$

    $$a(12b + 18) = 1 \quad \text{(Equation 1)}$$

  2. Set up the Second Equation from Given Info: We are given $P(X > 1, Y < 2) = 4/9$.

    • $X > 1 \implies X=2$.
    • $Y < 2 \implies Y=0, 1$. $$P(X=2, Y=0) + P(X=2, Y=1) = \frac{4}{9}$$ $$f_{XY}(2,0) + f_{XY}(2,1) = \frac{4}{9}$$ $$a(b(2)+0) + a(b(2)+1) = \frac{4}{9}$$ $$a(2b) + a(2b+1) = \frac{4}{9}$$ $$a(4b+1) = \frac{4}{9} \quad \text{(Equation 2)}$$

  3. Solve the System of Equations for a and b: From Equation 2, we have $a = \frac{4}{9(4b+1)}$. Substitute this into Equation 1:

    $$\frac{4}{9(4b+1)}(12b+18) = 1$$

    $$4(12b+18) = 9(4b+1)$$

    $$48b + 72 = 36b + 9$$

    $$12b = -63 \implies b = -\frac{63}{12} = -\frac{21}{4}$$

    Now find a using Equation 2:

    $$a(4(-\frac{21}{4})+1) = \frac{4}{9} \implies a(-21+1) = \frac{4}{9} \implies a(-20) = \frac{4}{9} \implies a = -\frac{4}{180} = -\frac{1}{45}$$

  4. Find the Required Value: The question asks for $f_{XY}(2,0)$.

    $$f_{XY}(2,0) = a(b(2)+0) = 2ab$$

    $$f_{XY}(2,0) = 2 \left(-\frac{1}{45}\right) \left(-\frac{21}{4}\right) = \frac{42}{180} = \frac{7}{30}$$

    As a decimal, this is approximately 0.2333… Rounding to two decimal places gives 0.23.

Question 6: Conditional Probability Logic

Problem: A fair coin is tossed 7 times. Let $X$ be the total number of heads and $Y$ be the number of heads before the first tail. (If there is no tail in all the seven tosses, then $Y=7$). What is the value of $f_{Y|X=x}(0|0)$? Write your answer correct to two decimal places.

Answer: 1.00

Concepts Explained 💡

  • Conditional Probability: The question asks for $f_{Y|X}(0|0)$, which is the probability $P(Y=0 | X=0)$. This is the probability that event $Y=0$ occurs, given that we know for a fact that event $X=0$ has occurred.
  • Logical Implication: If the occurrence of event B guarantees the occurrence of event A, then the conditional probability $P(A|B)$ is 1.

Step-by-Step Solution

  1. Understand the Given Event: The condition is $X=0$.

    • $X$ is the total number of heads in 7 tosses.
    • $X=0$ means there were zero heads.
    • The only possible outcome for the 7 tosses is TTTTTTT.

  2. Understand the Event We’re Finding the Probability Of: We want the probability of $Y=0$.

    • $Y$ is the number of heads before the first tail.
    • $Y=0$ means there were zero heads before the first tail occurred. This implies that the very first toss must be a Tail.

  3. Combine the Logic:

    • We are given that the outcome was TTTTTTT (because $X=0$).
    • In the outcome TTTTTTT, is it true that the number of heads before the first tail is zero?
    • Yes. The first toss is a Tail. There are no heads before it. Therefore, $Y=0$.

  4. Conclusion: The event $X=0$ (outcome is TTTTTTT) logically implies that the event $Y=0$ must also have occurred. When a given condition guarantees the outcome, the conditional probability is 1.

    $$P(Y=0 | X=0) = 1$$

    The answer correct to two decimal places is 1.00.

Question 7: Combinations for a Joint PMF

Problem: From a group of 3 members of party A, 2 members of party B, and 1 member of party C, a committee of two people is to be selected uniformly at random. Let $X$ denote the number of party A members and $Y$ denote the number of party B members on the committee. Find the value of $f_{XY}(1, 1)$. Write your answer correct to two decimal places.

Answer: 0.40

Concepts Explained 💡

  • Combinations: The number of ways to choose a subset of items from a larger set without regard to the order of selection. The formula is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
  • Classical Probability: When all outcomes are equally likely, the probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.

Step-by-Step Solution

  1. Identify the Goal: We need to find $f_{XY}(1, 1)$, which is the probability $P(X=1, Y=1)$.

    • $X=1$ means the committee has 1 member from party A.
    • $Y=1$ means the committee has 1 member from party B.
    • Since the committee size is 2, this means the committee must consist of exactly one person from A and one from B.

  2. Calculate the Number of Favorable Outcomes: We need to choose 1 person from A AND 1 person from B.

    • Ways to choose 1 from 3 party A members: $\binom{3}{1} = 3$.
    • Ways to choose 1 from 2 party B members: $\binom{2}{1} = 2$.
    • Total favorable outcomes = $3 \times 2 = 6$.

  3. Calculate the Total Number of Possible Outcomes: We are choosing a committee of 2 people from a total group of $3+2+1=6$ people.

    • Total ways to choose 2 from 6: $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$.

  4. Calculate the Probability:

    $$P(X=1, Y=1) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{6}{15} = \frac{2}{5}$$

    As a decimal, this is 0.40.

Question 8: Multinomial Probability

Problem: Suppose a fair six-sided die is rolled five times independently. Let $X$ denote the number of times a 1 is obtained and let $Y$ denote the number of times a 6 is obtained. Find $Pr(X = 1, Y = 1)$. Enter the answer correct to three decimal places.

Answer: 0.165

Concepts Explained 💡

  • Multinomial Distribution: An extension of the binomial distribution. It describes the probability of observing a specific count for each of several categories in a fixed number of independent trials ($n$), where each trial has the same probability for each category.
  • Multinomial PMF: The formula is: $$P(X_1=k_1, ..., X_m=k_m) = \frac{n!}{k_1!k_2!...k_m!} p_1^{k_1} p_2^{k_2} ... p_m^{k_m}$$ where $k_i$ is the number of times outcome $i$ occurs, and $p_i$ is the probability of outcome $i$ in a single trial.

Step-by-Step Solution

  1. Define the Experiment and Outcomes:

    • Number of trials ($n$) = 5 (five rolls of a die).
    • We are interested in three categories of outcomes for each roll:
      1. Getting a ‘1’: Probability $p_1 = 1/6$. We want this to happen once ($k_1=1$).
      2. Getting a ‘6’: Probability $p_2 = 1/6$. We want this to happen once ($k_2=1$).
      3. Getting anything else (2, 3, 4, or 5): Probability $p_3 = 4/6 = 2/3$. This must happen the remaining times: $k_3 = 5 - 1 - 1 = 3$.

  2. Apply the Multinomial Formula: We need to calculate $P(X=1, Y=1)$.

    $$P(X=1, Y=1) = \frac{5!}{1! \cdot 1! \cdot 3!} \left(\frac{1}{6}\right)^1 \left(\frac{1}{6}\right)^1 \left(\frac{4}{6}\right)^3$$

  3. Calculate the Components:

    • Combinatorial coefficient: $\frac{5!}{1!1!3!} = \frac{120}{1 \cdot 1 \cdot 6} = 20$. This represents the number of ways to arrange the outcomes (e.g., 1, 6, Other, Other, Other).
    • Probabilities: $\left(\frac{1}{6}\right)^1 \left(\frac{1}{6}\right)^1 \left(\frac{4}{6}\right)^3 = \frac{1}{6} \times \frac{1}{6} \times \frac{64}{216} = \frac{64}{7776}$.

  4. Compute the Final Probability:

    $$P(X=1, Y=1) = 20 \times \frac{64}{7776} = \frac{1280}{7776}$$

    Simplifying the fraction gives $\frac{40}{243}$.

    $$ \frac{40}{243} \approx 0.164609...$$

    Rounding to three decimal places gives 0.165.

Questions 9 & 10: Marginal PMFs from a Uniform Distribution

Problem: Let random variables $X$ and $Y$ be uniformly distributed over the set of integers $x$ and $y$ satisfying $-1 \le y \le 1$ and $-2 \le x+y \le 2$. 9) Find the marginal PMF of $X$. 10) Find the marginal PMF of $Y$.

Answer:

9) The correct marginal PMF for $X$ is the second option: $f_X(x) = \begin{cases} 1/15 & \text{if } x = -3, 3 \ 2/15 & \text{if } x = -2, 2 \ 3/15 & \text{if } x = -1, 0, 1 \ 0, & \text{otherwise} \end{cases}$

10) The correct marginal PMF for $Y$ is the second option: $f_Y(y) = \begin{cases} 1/3, & \text{if } y = -1, 0, 1 \ 0, & \text{otherwise} \end{cases}$

Concepts Explained 💡

  • Discrete Uniform Distribution: A distribution where all possible outcomes are equally likely. If there are $N$ outcomes, the probability of any single outcome is $1/N$.
  • Finding the Sample Space: The first step is to carefully list all pairs of integers $(x, y)$ that satisfy the given constraints.
  • Marginal PMF: As seen before, to find the marginal PMF of one variable, you sum the joint probabilities over all possible values of the other variable.

Step-by-Step Solution

  1. List all valid $(x, y)$ pairs (the sample space): The constraints are $y \in {-1, 0, 1}$ and $-2-y \le x \le 2-y$.

    • For y = -1: $-1 \le x \le 3 \implies$ Pairs are $(-1,-1), (0,-1), (1,-1), (2,-1), (3,-1)$. (5 pairs)
    • For y = 0: $-2 \le x \le 2 \implies$ Pairs are $(-2,0), (-1,0), (0,0), (1,0), (2,0)$. (5 pairs)
    • For y = 1: $-3 \le x \le 1 \implies$ Pairs are $(-3,1), (-2,1), (-1,1), (0,1), (1,1)$. (5 pairs)

  2. Determine the Joint Probability: There are a total of $5+5+5=15$ possible pairs. Since the distribution is uniform, each pair has a probability of $f_{XY}(x,y) = 1/15$.

  3. Calculate the Marginal PMF of X ($f_X(x)$): Sum the probabilities for each value of $x$.

    • $f_X(-3) = P(-3,1) = 1/15$
    • $f_X(-2) = P(-2,0) + P(-2,1) = 1/15 + 1/15 = 2/15$
    • $f_X(-1) = P(-1,-1) + P(-1,0) + P(-1,1) = 3/15$
    • $f_X(0) = P(0,-1) + P(0,0) + P(0,1) = 3/15$
    • $f_X(1) = P(1,-1) + P(1,0) + P(1,1) = 3/15$
    • $f_X(2) = P(2,-1) + P(2,0) = 2/15$
    • $f_X(3) = P(3,-1) = 1/15$ This matches the second option for question 9.

  4. Calculate the Marginal PMF of Y ($f_Y(y)$): Sum the probabilities for each value of $y$.

    • $f_Y(-1)$: There are 5 pairs where $y=-1$. Sum = $5 \times (1/15) = 5/15 = 1/3$.
    • $f_Y(0)$: There are 5 pairs where $y=0$. Sum = $5 \times (1/15) = 5/15 = 1/3$.
    • $f_Y(1)$: There are 5 pairs where $y=1$. Sum = $5 \times (1/15) = 5/15 = 1/3$. This matches the second option for question 10.

Questions 11 & 12: Joint Distributions from Placement Problems

Problem: A teacher places three candies into three boxes, with each candy independently and uniformly likely to go into any box. Let $X_i$ be the number of candies in Box $i$, and let $N$ be the number of occupied boxes. 11) Obtain the joint distribution of $(X_1, N)$. 12) Obtain the joint distribution of $(X_1, X_2)$.

Answer:

11) The correct joint distribution for $(X_1, N)$ is option (b) from image image_309a1b.png. 12) The correct joint distribution for $(X_1, X_2)$ is option (a) from image image_309db7.png.

Concepts Explained 💡

  • Sample Space Enumeration: For problems with a small number of trials and outcomes, we can list all possibilities to find probabilities.
  • Multinomial Coefficients: As seen in Q8, this helps count the number of ways to distribute distinct items into distinct bins. The number of ways to place $n$ items into $m$ bins, with $k_i$ items in bin $i$, is $\frac{n!}{k_1!k_2!…k_m!}$.
  • Joint Distribution Table: A table summarizing the probabilities $P(X=x, Y=y)$ for all possible pairs $(x, y)$.

Step-by-Step Solution

  1. Total Outcomes: Each of the 3 candies can go into any of the 3 boxes, so the total number of equally likely outcomes is $3 \times 3 \times 3 = 27$.

  2. Enumerate Configurations and Count Ways: A configuration is a tuple $(x_1, x_2, x_3)$ where $x_i$ is the number of candies in Box $i$.

    • Type (3,0,0): All 3 candies in one box.
      • Configurations: (3,0,0), (0,3,0), (0,0,3).
      • Ways for (3,0,0): $\binom{3}{3}=1$. All candies go to Box 1.
      • Total ways for this type: 3 configurations $\times$ 1 way each = 3.
    • Type (2,1,0): 2 in one box, 1 in another.
      • Configurations: (2,1,0), (2,0,1), (1,2,0), (0,2,1), (1,0,2), (0,1,2). (6 total)
      • Ways for (2,1,0): Choose 2 of 3 candies for Box 1: $\binom{3}{2}=3$.
      • Total ways for this type: 6 configurations $\times$ 3 ways each = 18.
    • Type (1,1,1): 1 candy in each box.
      • Configuration: (1,1,1).
      • Ways for (1,1,1): The 3 distinct candies can be arranged in $3! = 6$ ways.
      • Total ways for this type: 1 configuration $\times$ 6 ways = 6.
    • Check: $3 + 18 + 6 = 27$. Correct.

  3. Construct the Joint Table for $(X_1, N)$ (Question 11): For each configuration $(x_1, x_2, x_3)$ with its associated number of ways, find the corresponding $(X_1, N)$ pair.

    • $P(X_1=3, N=1)$: From config (3,0,0). Ways = 1. Prob = 1/27.
    • $P(X_1=2, N=2)$: From configs (2,1,0) & (2,0,1). Ways = 3+3 = 6. Prob = 6/27.
    • $P(X_1=1, N=2)$: From configs (1,2,0) & (1,0,2). Ways = 3+3 = 6. Prob = 6/27.
    • $P(X_1=1, N=3)$: From config (1,1,1). Ways = 6. Prob = 6/27.
    • $P(X_1=0, N=1)$: From configs (0,3,0) & (0,0,3). Ways = 1+1 = 2. Prob = 2/27.
    • $P(X_1=0, N=2)$: From configs (0,2,1) & (0,1,2). Ways = 3+3 = 6. Prob = 6/27.
    • All other probabilities are 0. This matches table (b).

    Final Table for (X₁, N):

    X₁ \ N123
    02/276/270
    106/276/27
    206/270
    31/2700

  4. Construct the Joint Table for $(X_1, X_2)$ (Question 12): For each configuration $(x_1, x_2, x_3)$ with its ways, find the $(X_1, X_2)$ pair.

    • $P(X_1=3, X_2=0)$: From config (3,0,0). Ways=1. Prob = 1/27.
    • $P(X_1=2, X_2=1)$: From config (2,1,0). Ways=3. Prob = 3/27.
    • $P(X_1=2, X_2=0)$: From config (2,0,1). Ways=3. Prob = 3/27.
    • $P(X_1=1, X_2=2)$: From config (1,2,0). Ways=3. Prob = 3/27.
    • $P(X_1=1, X_2=1)$: From config (1,1,1). Ways=6. Prob = 6/27.
    • $P(X_1=1, X_2=0)$: From config (1,0,2). Ways=3. Prob = 3/27.
    • …and so on for all configs. This matches table (a).

    Final Table for (X₁, X₂):

    X₁ \ X₂0123
    01/273/273/271/27
    13/276/273/270
    23/273/2700
    31/27000

Joint PMF of two discrete random variables