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.