Question 1205135
**1. Determine the Total Number of Possible Combinations**

* Since there are three characters on each bottlecap and each character can be repeated, the total number of possible combinations is: 
   * 3 characters * 3 characters * 3 characters = 27 possible combinations

**2. Calculate the Probability of Winning with One Bottlecap**

* There are 5 winning combinations.
* The probability of winning with a single bottlecap is: 
   * 5 winning combinations / 27 total combinations = 5/27 

**3. Calculate the Probability of Not Winning with One Bottlecap**

* The probability of not winning with a single bottlecap is: 
   * 1 - (5/27) = 22/27

**4. Calculate the Number of Bottles for Student B to Have a 1% Chance of Winning**

* Let 'n' be the number of bottles Student B needs to buy.
* The probability of not winning with 'n' bottles is (22/27)^n

* We want to find 'n' such that the probability of not winning is 99% (1 - 1% = 0.99): 
   (22/27)^n = 0.99

* Solve for 'n':
   n = log(0.99) / log(22/27) 
   n ≈ 1.23

* Since Student B cannot buy a fraction of a bottle, he needs to buy **at least 2 bottles** to have a 1% chance of winning.

**5. Calculate the Probability of Winning for Student C with 24 Bottlecaps**

* The probability of not winning with 24 bottlecaps is (22/27)^24 ≈ 0.0068

* The probability of winning with 24 bottlecaps is 1 - 0.0068 = 0.9932

* **Student C has a 99.32% chance of winning with 24 bottlecaps.**

**Note:**

* This calculation assumes that each bottlecap is an independent event. 
* In reality, the probability of winning might slightly decrease as more bottles are opened, as the number of remaining winning combinations would decrease. However, the difference would be negligible in this scenario.

Let me know if you have any other questions!