If 3 winning tickets are drawn from an urn of 100 tickets, then what is the probability of winning a person who buys 4 tickets?

To calculate the probability of winning when a person buys 4 tickets out of 100, with 3 winning tickets drawn, we can approach this using combinations and the concept of complementary probability.

### Step 1: Define the problem.
- Total number of tickets: \( 100 \)
- Number of tickets bought by the person: \( 4 \)
- Number of winning tickets: \( 3 \)

### Step 2: Calculate the total number of ways to choose 3 winning tickets from 100.
The total number of ways to draw 3 winning tickets from 100 is given by the combination formula \( C(n, k) \):
\[
\text{Total combinations} = C(100, 3) = \frac{100!}{3!(100-3)!}
\]

### Step 3: Calculate the number of favorable outcomes where the person wins.
There are different scenarios in which the person wins:

- The person gets exactly 1 winning ticket out of their 4.
- The person gets exactly 2 winning tickets out of their 4.
- The person gets exactly 3 winning tickets out of their 4.

Let’s calculate each:

1. **Exactly 1 winning ticket:**
   \[
   C(4, 1) \times C(96, 2)
   \]
   Here, \( C(4, 1) \) represents the ways to choose 1 winning ticket from the 4 tickets bought, and \( C(96, 2) \) represents the ways to choose the remaining 2 winning tickets from the 96 other tickets.

2. **Exactly 2 winning tickets:**
   \[
   C(4, 2) \times C(96, 1)
   \]
   Here, \( C(4, 2) \) represents the ways to choose 2 winning tickets from the 4 tickets bought, and \( C(96, 1) \) represents the ways to choose the remaining 1 winning ticket from the 96 other tickets.

3. **Exactly 3 winning tickets:**
   \[
   C(4, 3) \times C(96, 0)
   \]
   Here, \( C(4, 3) \) represents the ways to choose 3 winning tickets from the 4 tickets bought, and \( C(96, 0) \) is simply 1 because there are no more tickets to choose from.

### Step 4: Sum the favorable outcomes.
Now we sum the favorable outcomes from each scenario:
\[
\text{Favorable outcomes} = \left[ C(4, 1) \times C(96, 2) \right] + \left[ C(4, 2) \times C(96, 1) \right] + \left[ C(4, 3) \times C(96, 0) \right]
\]

### Step 5: Calculate the probability.
The probability of winning is the ratio of the favorable outcomes to the total possible outcomes:
\[
\text{Probability of winning} = \frac{\text{Favorable outcomes}}{C(100, 3)}
\]

Let me compute this for you:

1. \( C(100, 3) = \frac{100 \times 99 \times 98}{3 \times 2 \times 1} = 161700 \)
2. \( C(4, 1) = 4 \), \( C(96, 2) = \frac{96 \times 95}{2} = 4560 \)
3. \( C(4, 2) = 6 \), \( C(96, 1) = 96 \)
4. \( C(4, 3) = 4 \), \( C(96, 0) = 1 \)

Thus, the favorable outcomes:
\[
(4 \times 4560) + (6 \times 96) + (4 \times 1) = 18240 + 576 + 4 = 18820
\]

Finally, the probability:
\[
\text{Probability of winning} = \frac{18820}{161700} \approx 0.1164
\]

So, the probability of winning is approximately **0.1164**, or **11.64%**.

Would you like further details or have any questions?

### Related Questions:
1. How would the probability change if the number of winning tickets was increased to 5?
2. What is the probability of not winning any prize if a person buys 4 tickets?
3. How does buying more tickets affect the probability of winning?
4. What would be the probability if only 2 tickets were purchased instead of 4?
5. How does the total number of tickets affect the winning probability?

### Tip:
When dealing with probabilities involving combinations, it’s useful to think about complementary probabilities and use combination formulas to simplify calculations.

Learn how to calculate the probability of winning when buying 4 tickets out of 100, with 3 winning tickets drawn. This problem involves combinatorics and probability theory, covering the combination formula and complementary probability. Suitable for students and enthusiasts interested in advanced probability calculations.

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