Problem 8 - Applied Statistics and Probability for Engineers

Question

Suppose that the random variables \(X, Y,\) and \(Z\) have the following joint probability distribution. $$ \begin{array}{|cc|c|c|} \hline x & y & z & f(x, y, z) \\ \hline 1 & 1 & 1 & 0.05 \\ \hline 1 & 1 & 2 & 0.10 \\ \hline 1 & 2 & 1 & 0.15 \\ \hline 1 & 2 & 2 & 0.20 \\ \hline 2 & 1 & 1 & 0.20 \\ \hline 2 & 1 & 2 & 0.15 \\ \hline 2 & 2 & 1 & 0.10 \\ \hline 2 & 2 & 2 & 0.05 \\ \hline \end{array} $$ Determine the following: (a) \(P(X=2)\) (b) \(P(X=1, Y=2)\) (c) \(P(Z<1.5)\) (d) \(P(X=1 \quad\) or \(\quad Z=2)\) (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 \(Z=2\)

Step-by-Step Solution

Answer

(a) 0.50, (b) 0.35, (c) 0.50, (d) 0.70, (e) 1.50, (f) 0.30, (g) 0.20, (h) 0.40, (i) P(X=1|Y=1,Z=2)=0.40, P(X=2|Y=1,Z=2)=0.60.

Step 1: Find P(X=2)

We need to sum the probabilities where \(X\) equals 2. From the table:\[P(X=2) = f(2,1,1) + f(2,1,2) + f(2,2,1) + f(2,2,2) = 0.20 + 0.15 + 0.10 + 0.05 = 0.50\]

Step 2: Find P(X=1, Y=2)

We sum the probabilities where \(X=1\) and \(Y=2\):\[P(X=1, Y=2) = f(1,2,1) + f(1,2,2) = 0.15 + 0.20 = 0.35\]

Step 3: Find P(Z<1.5)

\(Z\) can only be 1 or 2. We sum probabilities where \(Z=1\):\[P(Z<1.5) = f(1,1,1) + f(1,2,1) + f(2,1,1) + f(2,2,1) = 0.05 + 0.15 + 0.20 + 0.10 = 0.50\]

Step 4: Find P(X=1 or Z=2)

We calculate \(P(X=1)\), \(P(Z=2)\), and \(P(X=1,Z=2)\) to use the inclusion-exclusion principle:\[P(X=1) = 0.05 + 0.10 + 0.15 + 0.20 = 0.50\]\[P(Z=2) = 0.10 + 0.20 + 0.15 + 0.05 = 0.50\]\[P(X=1, Z=2) = 0.10 + 0.20 = 0.30\]Then,\[P(X=1 \text{ or } Z=2) = P(X=1) + P(Z=2) - P(X=1, Z=2) = 0.50 + 0.50 - 0.30 = 0.70\]

Step 5: Calculate E(X)

The expected value \(E(X)\) is calculated as follows:\[E(X) = 1 \cdot P(X=1) + 2 \cdot P(X=2) = 1 \times 0.50 + 2 \times 0.50 = 1.50\]

Step 6: Compute P(X=1 | Y=1)

Conditional probability is given by\[P(X=1 \mid Y=1) = \frac{P(X=1, Y=1)}{P(Y=1)}\]Calculate each:\[P(X=1, Y=1) = f(1,1,1) + f(1,1,2) = 0.05 + 0.10 = 0.15\]\[P(Y=1) = 0.05 + 0.10 + 0.20 + 0.15 = 0.50\]Thus,\[P(X=1 \mid Y=1) = \frac{0.15}{0.50} = 0.30\]

Step 7: Find P(X=1, Y=1 | Z=2)

\[P(X=1, Y=1 \mid Z=2) = \frac{P(X=1, Y=1, Z=2)}{P(Z=2)}\]Calculate each:\[P(X=1, Y=1, Z=2) = f(1,1,2) = 0.10\]\[P(Z=2) = 0.50\]Thus,\[P(X=1, Y=1 \mid Z=2) = \frac{0.10}{0.50} = 0.20\]

Step 8: Find P(X=1 | Y=1, Z=2)

\[P(X=1 \mid Y=1, Z=2) = \frac{P(X=1, Y=1, Z=2)}{P(Y=1, Z=2)}\]We earlier found \(P(X=1, Y=1, Z=2)=0.10\).Calculate \(P(Y=1, Z=2)\):\[P(Y=1, Z=2) = f(1,1,2) + f(2,1,2) = 0.10 + 0.15 = 0.25\]Thus, \[P(X=1 \mid Y=1, Z=2) = \frac{0.10}{0.25} = 0.40\]

Step 9: Conditional distribution of X given Y=1, Z=2

For the conditional distribution \(P(X \mid Y=1, Z=2)\), we find the probabilities for each possible value of \(X\) given \(Y=1\) and \(Z=2\):\[P(X=1 \mid Y=1, Z=2) = 0.40\]\[P(X=2 \mid Y=1, Z=2) = 1 - P(X=1 \mid Y=1, Z=2) = 0.60\]

Key Concepts

Conditional Probability

Conditional probability allows us to calculate the likelihood of an event occurring based on the occurrence of another related event.
It's a way to answer questions like, "What's the probability of X happening if we know that Y has happened?"
In mathematical terms, conditional probability is expressed as \(P(A|B)\), which can be read as "the probability of A given B."
Here, \(P(A|B)\) is calculated as:

• \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\)

This formula shows that to find the probability of A occurring given B, we divide the probability of both A and B occurring by the probability of just B occurring.
For instance, calculating \(P(X=1 \mid Y=1)\) involves determining the probability that both X=1 and Y=1 happen and dividing that by the total probability of Y=1 happening.
This concept is essential for decision-making and predictions in uncertain environments.

Expected Value

The expected value is a crucial concept in probability, representing the average or mean value that you'd expect if you could repeat an experiment infinite times.
Defined as \(E(X)\), the expected value provides a measure of the center of a random variable's distribution.
It's calculated by taking each possible value of the random variable, multiplying it by its probability, and summing all these products:

• \(E(X) = \sum (x_i \cdot P(x_i))\)

For example, if you want to find the expected value of X in a probability distribution, like the one provided in an exercise, you'd multiply each value of X by its respective probability and add them together.
This measurement helps in decision-making processes by providing an understanding of what can be expected on average in the long run.

Random Variables

Random variables are essential elements of probability theory.
They are variables that take on different numerical values, each associated with a probability under the uncertainty of a particular experiment or process.
Random variables can be discrete or continuous:

• Discrete random variables have distinct, separate values (e.g., number of heads in 10 coin flips).
• Continuous random variables can take any value within a range (e.g., time taken for an event to happen).

In exercises involving random variables, such as the one outlined, they are used to model uncertainties and compute various probabilities and expected outcomes.
They provide the foundation for integrating probability into real-world circumstances, enabling predictions and analysis of complex systems.

Probability Distribution

A probability distribution gives a complete description of a random variable by specifying the probabilities associated with each of its possible values.
It characterizes the way in which probabilities are spread over the values of the random variable.
Such distributions can be presented using probability mass functions for discrete variables or probability density functions for continuous variables.
In the example exercise, the joint probability distribution for \(X\), \(Y\), and \(Z\) assigns a probability to each combination of these variables.
The table essentially shows the probability mass function, which gives specific probabilities for combinations of \(X, Y, Z\) values.
Understanding these distributions is key to interpreting and working with the likelihood of different outcomes in probabilistic scenarios.