Question
The Pellard High School football team is holding a raffle to raise money for new uniforms. They sold a total of 500 raffle tickets for the Grand Prize. • If one ticket is purchased, what is the probability of winning? • How many tickets must an individual buy to have a 10% chance of winning? • A computer simulation is used to model the 500 tickets in the raffle. Noah’s name is entered into the simulation 5 times. The simulation is run 500 times. He determines that his name should be selected in exactly 1% of the simulations. Is he correct? Enter your answers and your work or explanation in the box provided.
Answers
Answered by
GPT-4o mini
To solve the problems step by step:
1. **Probability of winning with one ticket:**
- There are 500 raffle tickets sold, and if you buy 1 ticket, the probability of winning (P) is given by:
\[
P(\text{winning with 1 ticket}) = \frac{\text{Number of winning tickets}}{\text{Total number of tickets}} = \frac{1}{500} = 0.002
\]
- Thus, the probability of winning with one ticket is **0.002** or **0.2%**.
2. **Number of tickets needed for a 10% chance of winning:**
- To find the number of tickets (n) needed for at least a 10% chance of winning, we can use the complementary probability (the probability of not winning) and set it up as follows:
\[
P(\text{not winning with n tickets}) = \left(1 - \frac{1}{500}\right)^n
\]
- We want the probability of winning to be at least 10%, which is:
\[
P(\text{winning}) = 1 - P(\text{not winning}) \geq 0.1
\]
- Therefore:
\[
1 - \left(1 - \frac{1}{500}\right)^n \geq 0.1
\]
- Rearranging yields:
\[
\left(1 - \frac{1}{500}\right)^n \leq 0.9
\]
- Taking the logarithm of both sides, we get:
\[
n \cdot \ln\left(1 - \frac{1}{500}\right) \leq \ln(0.9)
\]
- Since \(\ln\left(1 - \frac{1}{500}\right)\) is a negative number, we can switch the inequality when dividing:
\[
n \geq \frac{\ln(0.9)}{\ln\left(1 - \frac{1}{500}\right)}
\]
- Calculating this gives:
\[
\ln(0.9) \approx -0.10536
\]
\[
\ln\left(1 - \frac{1}{500}\right) \approx -\frac{1}{500} \approx -0.002
\]
- Thus:
\[
n \geq \frac{-0.10536}{-0.002} \approx 52.68
\]
- Rounding up, an individual must buy **53 tickets** to have at least a 10% chance of winning.
3. **Noah's name entered into the computer simulation:**
- Noah's name is entered into the simulation 5 times out of 500.
- The expected probability of winning for Noah is:
\[
P(\text{Noah winning}) = \frac{5}{500} = 0.01
\]
- The simulation runs 500 times, and if it is said that Noah’s name should be selected in exactly 1% of the simulations, this means:
- In terms of expected outcomes:
\[
500 \cdot 0.01 = 5 \text{ times}
\]
- So, Noah should expect to win about 5 times in 500 simulations based on his entries, and since the simulation returns 1% of the time, Noah’s understanding is correct.
In summary, the solutions are:
- Probability of winning with 1 ticket: **0.002**
- Tickets needed for a 10% chance: **53 tickets**
- Noah's assessment is correct regarding his expected win rate in the simulation.
1. **Probability of winning with one ticket:**
- There are 500 raffle tickets sold, and if you buy 1 ticket, the probability of winning (P) is given by:
\[
P(\text{winning with 1 ticket}) = \frac{\text{Number of winning tickets}}{\text{Total number of tickets}} = \frac{1}{500} = 0.002
\]
- Thus, the probability of winning with one ticket is **0.002** or **0.2%**.
2. **Number of tickets needed for a 10% chance of winning:**
- To find the number of tickets (n) needed for at least a 10% chance of winning, we can use the complementary probability (the probability of not winning) and set it up as follows:
\[
P(\text{not winning with n tickets}) = \left(1 - \frac{1}{500}\right)^n
\]
- We want the probability of winning to be at least 10%, which is:
\[
P(\text{winning}) = 1 - P(\text{not winning}) \geq 0.1
\]
- Therefore:
\[
1 - \left(1 - \frac{1}{500}\right)^n \geq 0.1
\]
- Rearranging yields:
\[
\left(1 - \frac{1}{500}\right)^n \leq 0.9
\]
- Taking the logarithm of both sides, we get:
\[
n \cdot \ln\left(1 - \frac{1}{500}\right) \leq \ln(0.9)
\]
- Since \(\ln\left(1 - \frac{1}{500}\right)\) is a negative number, we can switch the inequality when dividing:
\[
n \geq \frac{\ln(0.9)}{\ln\left(1 - \frac{1}{500}\right)}
\]
- Calculating this gives:
\[
\ln(0.9) \approx -0.10536
\]
\[
\ln\left(1 - \frac{1}{500}\right) \approx -\frac{1}{500} \approx -0.002
\]
- Thus:
\[
n \geq \frac{-0.10536}{-0.002} \approx 52.68
\]
- Rounding up, an individual must buy **53 tickets** to have at least a 10% chance of winning.
3. **Noah's name entered into the computer simulation:**
- Noah's name is entered into the simulation 5 times out of 500.
- The expected probability of winning for Noah is:
\[
P(\text{Noah winning}) = \frac{5}{500} = 0.01
\]
- The simulation runs 500 times, and if it is said that Noah’s name should be selected in exactly 1% of the simulations, this means:
- In terms of expected outcomes:
\[
500 \cdot 0.01 = 5 \text{ times}
\]
- So, Noah should expect to win about 5 times in 500 simulations based on his entries, and since the simulation returns 1% of the time, Noah’s understanding is correct.
In summary, the solutions are:
- Probability of winning with 1 ticket: **0.002**
- Tickets needed for a 10% chance: **53 tickets**
- Noah's assessment is correct regarding his expected win rate in the simulation.