Question 1209959
Let's break down this problem into two parts:

**Understanding the Problem**

* We have 15 seats in a row.
* 9 people are actors (A).
* 6 people are backstage workers (B).
* Tickets are assigned randomly.

**a) Probability an Actor and Backstage Worker Sit Together**

1. **Calculate the total number of arrangements:**
   The total number of ways to arrange the 15 people in the 15 seats is 15! (15 factorial).

2. **Calculate the number of arrangements where an actor and backstage worker sit together:**
   This is more complex. We'll use the complementary probability: calculate the arrangements where *no* actor and backstage worker sit together, and subtract it from the total.

   * **Consider pairs:**
     An actor and backstage worker sit together as a pair (AB or BA).

   * **Instead, calculate the probability that they are never together:**
     This calculation is very complex.

   * **Instead, a simpler approach is to use conditional probability:**
     * **Probability of the first seat being an actor:** 9/15
     * **Probability of the second seat being a backstage worker:** 6/14
     * **Probability of the first seat being a backstage worker:** 6/15
     * **Probability of the second seat being an actor:** 9/14

     * **Probability that the first two are an actor and backstage worker:** (9/15)*(6/14) + (6/15)*(9/14) = 54/210 + 54/210 = 108/210

    * **However, this only works for the first two seats. It is far more complicated to calculate for all seats.**

    * **Instead, we will use a simpler approach of approximation:**
     * Consider the number of pairs of seats. There are 14 pairs of adjacent seats.
     * The probability that any specific pair has an actor and backstage worker is approximately (9/15)(6/14) + (6/15)(9/14) = 108/210 = 18/35.
     * The probability that *at least one* pair has an actor and backstage worker is difficult to calculate precisely.
     * However, the probability that a specific pair has an actor and backstage worker is 18/35, so it is likely that they will sit together.

**b) Probability Actors Sit Together and Backstage Workers Sit Together**

1. **Treat actors as a single block:**
   Consider the 9 actors as one block (A) and the 6 backstage workers as another block (B).

2. **Arrange the blocks:**
   There are 2! (2 factorial) ways to arrange these two blocks (AB or BA).

3. **Arrange the actors:**
   There are 9! ways to arrange the 9 actors within their block.

4. **Arrange the backstage workers:**
   There are 6! ways to arrange the 6 backstage workers within their block.

5. **Calculate the number of favorable arrangements:**
   The number of favorable arrangements is 2! * 9! * 6!

6. **Calculate the total number of arrangements:**
   The total number of arrangements is 15!

7. **Calculate the probability:**
   Probability = (2! * 9! * 6!) / 15!

8. **Simplify:**
   Probability = (2 * 9! * 6!) / 15! = (2 * 6!) / (15 * 14 * 13 * 12 * 11 * 10) = (2 * 720) / 3603600 = 1440/3603600 = 1/2502.5

   Probability = 1440/3603600 = 4/10010

**Answers**

* **a) Probability an actor and backstage worker sit together:** This is a complex calculation and requires more advanced methods. It is likely a high probability.
* **b) Probability actors and backstage workers sit together:** 4/10010 or approximately 0.0003996.