91Ó°ÊÓ

Understanding the probability mass function (pmf) is essential when dealing with discrete random variables like the ones in this exercise. The pmf specifies the probability that a discrete random variable is exactly equal to some value. In our example, we have two random variables, \(X\) and \(Y\), each with their own pmfs.
The pmf for \(X\) indicates how likely it is to have 0, 1, or 2 headlights needing adjustment:

• \(p_X(0) = 0.5\)
• \(p_X(1) = 0.3\)
• \(p_X(2) = 0.2\)

The pmf for \(Y\) shows the probability distribution for the number of defective tires:

• \(p_Y(0) = 0.6\)
• \(p_Y(1) = 0.1\)
• \(p_Y(2) = 0.05\)
• \(p_Y(3) = 0.05\)
• \(p_Y(4) = 0.2\)

These functions help us understand the likelihood of each scenario, crucial for further calculations.

A key aspect of this exercise is that \(X\) and \(Y\) are independent random variables. Independence means that the occurrence of one event does not affect the probability of the other. This simplifies calculations since we can multiply the individual probabilities of each event.
For instance, in our problem, the probability that there are 0 headlights needing adjustment and 0 defective tires is simply the product of their individual probabilities:
\(p(0,0) = p_X(0) \times p_Y(0) = 0.5 \times 0.6 = 0.3\).

Recognizing independence allows us to construct a joint probability table efficiently because each joint probability is determined by multiplication of the simple pmfs. This principle is fundamental when dealing with multiple events simultaneously, as it enables straightforward computation of joint pmf.

The joint probability table is instrumental for visualizing the behavior of two independent random variables simultaneously. It lists the probabilities of each combination of \(X\) and \(Y\). Creating this table involves calculating joint probabilities for each pair and is particularly helpful in complex probability calculations.
For \(X\) values of 0, 1, and 2 and \(Y\) values of 0 through 4, each probability is calculated by multiplying the corresponding values from the two separate pmfs:

• For \((X=0, Y=0)\), we already computed \(p(0,0) = 0.3\).
• For \((X=1, Y=2)\), calculate: \(p(1,2) = p_X(1) \times p_Y(2) = 0.3 \times 0.05 = 0.015\).

This calculation is repeated for every combination, providing a comprehensive overview of joint probabilities. By interpreting this table, students can understand how individual events come together and learn to compute probabilities involving multiple variable outcomes.

Probability calculations involving independent random variables and joint probabilities require careful attention to detail. This is particularly true for events involving multiple conditions, such as \(P(X \leq 1, Y \leq 1)\), where both \(X\) and \(Y\) meet a specific criterion.
To compute such probabilities, sum all relevant joint probabilities from the table. For \(P(X \leq 1, Y \leq 1)\), consider:

• \(p(0, 0) = 0.3\)
• \(p(0, 1) = 0.05\)
• \(p(1, 0) = 0.18\)
• \(p(1, 1) = 0.03\)

Adding these gives \(P(X \leq 1, Y \leq 1) = 0.56\).
Additionally, verifying calculations using probability rules like independence ensures consistent results. Multiplying \(P(X \leq 1)\) and \(P(Y \leq 1)\) independently confirms this complex probability is correct, demonstrating the joint behavior of independent random variables.