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Suppose the random variables \(X, Y\), and \(Z\) have the following joint probability distribution. $$ \begin{array}{cccc} \hline x & y & z & f(x, y, z) \\ \hline 1 & 1 & 1 & 0.05 \\ 1 & 1 & 2 & 0.10 \\ 1 & 2 & 1 & 0.15 \\ 1 & 2 & 2 & 0.20 \\ 2 & 1 & 1 & 0.20 \\ 2 & 1 & 2 & 0.15 \\ 2 & 2 & 1 & 0.10 \\ 2 & 2 & 2 & 0.05 \\ \hline \end{array} $$ Determine the following: (a) \(P(X=2)\) (b) \(P(X=1, Y=2)\) (c) \(P(Z < 1.5)\) (d) \(P(X=1\) or \(Z=2)\) (e) \(E(X)\) (f) \(P(X=1 \mid Y=1)\) (g) \(P(X=1, Y=1 \mid Z=2)\) (h) \(P(X=1 \mid Y=1, Z=2)\) (i) Conditional probability distribution of \(X\) given that \(Y=1\) and \(Z=2\)

Short Answer

Expert verified
(a) 0.50; (b) 0.35; (c) 0.50; (d) 0.70; (e) 1.5; (f) 0.30; (g) 0.20; (h) 0.40; (i) P(X=1)=0.40, P(X=2)=0.60.

Step by step solution

01

Calculate P(X=2)

To find \(P(X=2)\), sum the probabilities where \(X=2\). These are all the entries in the joint distribution table where the first column is 2:\[P(X=2) = 0.20 + 0.15 + 0.10 + 0.05 = 0.50\]
02

Calculate P(X=1, Y=2)

To find \(P(X=1, Y=2)\), note that both \(X=1\) and \(Y=2\) must be true. These correspond to the sums of probabilities in the table where \(x=1\) and \(y=2\):\[P(X=1, Y=2) = 0.15 + 0.20 = 0.35\]
03

Calculate P(Z

The condition \(Z < 1.5\) implies \(Z=1\) (since probability values correspond only to integers in the table). Sum the probabilities where \(z=1\):\[P(Z<1.5) = 0.05 + 0.15 + 0.20 + 0.10 = 0.50\]
04

Calculate P(X=1 or Z=2)

Use the inclusion-exclusion principle. Find \(P(X=1)\), \(P(Z=2)\), and \(P(X=1, Z=2)\):\[P(X=1) = 0.05 + 0.10 + 0.15 + 0.20 = 0.50 \P(Z=2) = 0.10 + 0.20 + 0.15 + 0.05 = 0.50 \P(X=1, Z=2) = 0.10 + 0.20 = 0.30 \P(X=1 \text{ or } Z=2) = P(X=1) + P(Z=2) - P(X=1, Z=2) = 0.50 + 0.50 - 0.30 = 0.70\]
05

Calculate E(X)

To find the expected value \(E(X)\), multiply each \(x\)-value by its probability:\[E(X) = 1 \cdot 0.50 + 2 \cdot 0.50 = 1.5\]
06

Calculate P(X=1 | Y=1)

To find \(P(X=1 \mid Y=1)\), use the formula \(P(A | B) = \frac{P(A \cap B)}{P(B)}\). Here, \(A\) is \(X=1\), \(B\) is \(Y=1\):\[P(X=1 \cap Y=1) = 0.05 + 0.10 = 0.15 \P(Y=1) = 0.05 + 0.10 + 0.20 + 0.15 = 0.50 \P(X=1 \mid Y=1) = \frac{0.15}{0.50} = 0.30\]
07

Calculate P(X=1, Y=1 | Z=2)

Using conditional probability, find \(P(X=1, Y=1 \mid Z=2)\):\[P(X=1, Y=1, Z=2) = 0.10 \P(Z=2) = 0.50 \P(X=1, Y=1 \mid Z=2) = \frac{0.10}{0.50} = 0.20\]
08

Calculate P(X=1 | Y=1, Z=2)

Find \(P(X=1 \mid Y=1, Z=2)\) using the probabilities from the table:\[P(X=1, Y=1, Z=2) = 0.10 \P(Y=1, Z=2) = 0.10 + 0.15 = 0.25 \P(X=1 \mid Y=1, Z=2) = \frac{0.10}{0.25} = 0.40\]
09

Find Conditional Probability Distribution of X given Y=1 and Z=2

To find \(P(X=x \mid Y=1, Z=2)\) for each \(x\):- For \(x=1\):\[P(X=1 \mid Y=1, Z=2) = \frac{0.10}{0.25} = 0.40\]- For \(x=2\):\[P(X=2 \mid Y=1, Z=2) = \frac{0.15}{0.25} = 0.60\]Thus, the conditional distribution is \(P(X=1) = 0.40\) and \(P(X=2) = 0.60\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability
Conditional probability is a fundamental concept in probability theory. It refers to the probability of an event occurring, given that another event has already occurred. The notation is typically written as \(P(A|B)\), which reads as "the probability of \(A\) given \(B\)." A strong understanding of conditional probability helps in solving problems where events are dependent on each other.
For instance, in our exercise, when determining \(P(X=1 \mid Y=1)\), we want to know the probability of \(X=1\) happening when we already know that \(Y=1\) has occurred. This can be calculated using the formula:\[P(X=1 \mid Y=1) = \frac{P(X=1 \cap Y=1)}{P(Y=1)}\]The essential idea is narrowing down the sample space to only those outcomes where the condition \(Y=1\) is true. This makes our calculations more relevant and focused, given the occurrence of this condition.
Expected Value
The expected value is a crucial concept that provides a measure of the central tendency of a random variable. Essentially, it helps to understand the "average" outcome one can expect when experimenting with random variables. For a discrete random variable \(X\) with a probability mass function \(f(x)\), the expected value \(E(X)\) is calculated as follows:
  • Take each possible value of the random variable.
  • Multiply it by its probability.
  • Sum these products.
Mathematically, this is written as:\[E(X) = \sum x_i P(x_i)\]In the exercise provided, \(E(X)\) was computed by summing the products of each possible value of \(X\) (1 and 2) and their associated probabilities, resulting in \(1.5\).
Expected value allows you to anticipate the outcome over many trials or occurrences of a random event, which is particularly useful in fields like economics, finance, and decision-making processes.
Random Variables
Random variables are fundamentally used to quantify outcomes of random phenomena. They are variables that take on different values based on the outcome of a random event. These variables are categorized majorly into two types: discrete and continuous.
In our context, random variables \(X, Y,\) and \(Z\) can assume specific, countable outcomes (like 1 or 2 in this exercise). These are discrete random variables since they take on a finite number of values.
Understanding random variables involves recognizing their probability distributions, which provide the probabilities of each outcome the variable can take. In our joint probability distribution table, the random variables are interconnected, each with its probabilities tied to different combinations of \((X, Y, Z)\). Knowing how to interpret these distributions is key to calculating probabilities, expectations, and variances linked to these variables.
Probability Theory
Probability theory is the mathematical framework for describing random events and uncertainty. It forms the foundation for various advanced topics and applications in statistics and science.
At its core, probability theory deals with the likelihood of events occurring. It uses probability distributions and probability mass functions to quantify these likelihoods for random variables. For example, in the exercise, probability theory allows us to deduce outcomes like \(P(X=1 \text{ or } Z=2)\) using principles such as the inclusion-exclusion principle.
As a field, probability theory encompasses various sub-concepts including joint probabilities (considering two or more events simultaneously) and conditional probabilities (probabilities conditioned on other events' occurrences). Mastering probability theory is critical for understanding and building upon statistical concepts, designing experiments, and making informed decisions in the presence of uncertainty.

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Most popular questions from this chapter

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