SOLUTION: Student A was tasked to program an algorithm for a number generator. The generator will be used by Soda Co. for their bottlecaps. Soda Co. will be holding a bottlecap lottery w

Question 1205135: Student A was tasked to program an algorithm for a number generator. The generator will be used by Soda Co. for their bottlecaps.
Soda Co. will be holding a bottlecap lottery where 5 select bottlecaps will be winning combinations. There would be three characters printed on the bottlecaps with each character having the chance to be repeated. Those that are not winning codes are duplicated 10 times.

If Student B was to join, how many bottles of Soda will he need to buy to have a 1% chance at winning.
And if Student C would join too, what is the chance that having 24 bottlecaps have a chance at winning?
Answer by ElectricPavlov(122) (Show Source): You can put this solution on YOUR website!
**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!

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