https://file.aiquickdraw.com/mathbot/file/172641450788069bkkmcu.jpg

Let's solve the questions shown in the image step by step:

### Question 1: 
**If you buy 10 bars, what is the probability of winning at least one prize?**

This is a binomial probability problem. The probability of winning a prize in a single bar is \( p = \frac{1}{20} \), since a prize is wrapped in every 20th bar. 

The probability of **not winning** a prize in a single bar is:
\[
q = 1 - p = 1 - \frac{1}{20} = \frac{19}{20}
\]

The probability of **not winning any prize** in 10 bars is:
\[
q^{10} = \left(\frac{19}{20}\right)^{10}
\]

Thus, the probability of winning **at least one prize** is:
\[
1 - q^{10} = 1 - \left(\frac{19}{20}\right)^{10}
\]

Let's calculate that:

\[
q^{10} = \left(\frac{19}{20}\right)^{10} \approx 0.5987
\]
\[
P(\text{at least one prize}) = 1 - 0.5987 = 0.4013
\]

So, the probability of winning at least one prize is **0.401** (3 d.p.).

---

### Question 2:
**How many bars would you need to buy to be 99% sure of winning at least one prize?**

To be 99% sure of winning at least one prize, we want the probability of **not winning** any prize to be less than 1%. This is expressed as:
\[
q^n < 0.01
\]
\[
\left(\frac{19}{20}\right)^n < 0.01
\]

Taking the natural logarithm on both sides:
\[
n \ln\left(\frac{19}{20}\right) < \ln(0.01)
\]
\[
n < \frac{\ln(0.01)}{\ln\left(\frac{19}{20}\right)}
\]

Let’s calculate that:

\[
n = \frac{\ln(0.01)}{\ln\left(\frac{19}{20}\right)} \approx \frac{-4.6052}{-0.0513} \approx 89.79
\]

So, you would need to buy approximately **90 bars** to be 99% sure of winning at least one prize.

---

### Question 3:
**On an order for 10 bars placed during the promotion, what is the probability that Donna makes a loss?**

Let's calculate Donna's profit per bar:
- Donna buys each bar for £0.20.
- Donna sells each bar for £1.50.

If she gives away a £5 prize, she incurs an extra cost.

The total cost for 10 bars is:
\[
\text{Cost} = 10 \times 0.20 = £2
\]

If Donna **does not give away a prize**, her total revenue is:
\[
\text{Revenue} = 10 \times 1.50 = £15
\]
Profit if no prize is given:
\[
\text{Profit} = £15 - £2 = £13
\]

If Donna **gives away a prize**, her revenue is still £15, but her total cost increases to:
\[
\text{Cost} = £2 + £5 = £7
\]
Profit if a prize is given:
\[
\text{Profit} = £15 - £7 = £8
\]

So, Donna makes a loss if she gives away a prize. The probability of giving away a prize is the same as the probability of winning at least one prize in 10 bars, which we calculated in Question 1 to be 0.401.

Thus, the probability that Donna makes a loss is **0.401** (3 d.p.).

---

### Do you want more details or have any questions?

### Follow-up questions:
1. What would be the expected value of Donna's profit if she sells 10 bars?
2. How would the answers change if the promotion was changed to every 15th bar having a prize?
3. What is the probability of winning **exactly one prize** if you buy 10 bars?
4. How much would Donna need to charge per bar to ensure no loss on average, considering the current promotion?
5. How does the probability of winning at least one prize change as the number of bars increases?

### Tip:
In problems involving probabilities of multiple events, consider complementary events ("not happening") to simplify calculations of "at least one" scenarios.

Donna owns a failing chocolate shop. She wraps a £5 prize inside every 20th supersize bar of chocolate, mixes the bars, and sells them randomly. What is the probability of winning at least one prize if you buy 10 bars? How many bars would you need to buy to be 99% sure of winning at least one prize? What is the probability that Donna makes a loss during the promotion for 10 bars?

Explore the probability of winning a prize in a chocolate shop promotion. Learn to calculate the chance of winning at least one prize, how many chocolate bars you would need to buy to be 99% sure of winning, and how to assess profit and loss in a promotion. Suitable for students in Grades 10-12.

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