In this problem, we have three different bags, and they need to be loaded into three different vehicles (belonging to guests Josh, Stell, and Pablo). The staff must assign one bag to each guest, and the random variable $X$ represents the number of correct matches (i.e., the number of bags correctly assigned to their respective owners).

Step 1: Total Number of Possible Assignments

Since each bag can be assigned to one of the three guests, and no bags can be assigned more than once, the total number of ways the staff can assign the bags is:

$3! = 6$

These 6 assignments (permutations) are:

• (Josh, Stell, Pablo)
• (Josh, Pablo, Stell)
• (Stell, Josh, Pablo)
• (Stell, Pablo, Josh)
• (Pablo, Josh, Stell)
• (Pablo, Stell, Josh)

Step 2: Number of Correct Matches for Each Assignment

We now consider the number of correct matches in each assignment:

1. (Josh, Stell, Pablo): All bags are correctly matched. Thus, $X = 3$.
2. (Josh, Pablo, Stell): Only Josh has the correct bag. Thus, $X = 1$.
3. (Stell, Josh, Pablo): Only Stell has the correct bag. Thus, $X = 1$.
4. (Stell, Pablo, Josh): Only Pablo has the correct bag. Thus, $X = 1$.
5. (Pablo, Josh, Stell): Only Josh has the correct bag. Thus, $X = 1$.
6. (Pablo, Stell, Josh): No one has the correct bag. Thus, $X = 0$.

Step 3: Probability Distribution for $X$

Now, we calculate the probability of each value of $X$. We count the number of times each value of $X$ occurs and divide by the total number of possible assignments (which is 6):

• $X = 3$ occurs 1 time (out of 6).
• $X = 1$ occurs 4 times (out of 6).
• $X = 0$ occurs 1 time (out of 6).

Thus, the probability distribution of $X$ is:

$P(X = 0) = \frac{1}{6}, \quad P(X = 1) = \frac{4}{6}, \quad P(X = 3) = \frac{1}{6}$

Final Answer:

The probability distribution for $X$ is:

• $P(X = 0) = \frac{1}{6}$
• $P(X = 1) = \frac{4}{6}$
• $P(X = 3) = \frac{1}{6}$

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Would you like more details on how this probability distribution is derived, or do you have any other questions? Here are a few related questions you might find interesting:

1. What is the expected value of $X$ in this scenario?
2. How would the distribution change if there were more than three guests and bags?
3. What is the variance of the number of correct matches $X$?
4. How does the concept of derangements relate to this problem?
5. What would happen to the probability distribution if there were a rule about bag preferences?

Tip: When solving probability problems with permutations, it's helpful to count the total outcomes first and then classify them into different categories based on the number of correct matches or desired outcomes.