Five friends go to their favourite restaurant which has 8 different lunches on the menu. If the friends
are equally likely to pick any of the dinners, what is the probability that two or more of the friends will
order the same lunch?

To solve this problem, we can use the complementary probability approach. The idea is to first calculate the probability that all five friends choose different lunches and then subtract this from 1 to find the probability that two or more of the friends will order the same lunch.

### Step 1: Calculate the total number of possible outcomes
Each of the 5 friends can choose from 8 different lunches, so the total number of possible outcomes is:

\[
8^5
\]

### Step 2: Calculate the number of favorable outcomes (all friends choose different lunches)
To calculate the number of ways in which all 5 friends can choose different lunches, we use the concept of permutations because the order in which they choose the lunches matters. The first friend has 8 choices, the second friend has 7 remaining choices, the third friend has 6 choices, the fourth has 5, and the fifth has 4 choices. So, the number of favorable outcomes is:

\[
8 \times 7 \times 6 \times 5 \times 4
\]

### Step 3: Calculate the probability that all friends choose different lunches
The probability that all 5 friends choose different lunches is:

\[
P(\text{all different}) = \frac{8 \times 7 \times 6 \times 5 \times 4}{8^5}
\]

### Step 4: Calculate the probability that two or more friends choose the same lunch
The probability that two or more friends choose the same lunch is the complement of the probability that all friends choose different lunches:

\[
P(\text{two or more same}) = 1 - P(\text{all different})
\]

Let's calculate this:

\[
P(\text{all different}) = \frac{8 \times 7 \times 6 \times 5 \times 4}{8^5} = \frac{6720}{32768} \approx 0.205
\]

\[
P(\text{two or more same}) = 1 - 0.205 = 0.795
\]

So, the probability that two or more of the friends will order the same lunch is approximately **0.795** or **79.5%**.

Would you like more details on any part of this solution? 

Here are some related questions:

1. How would the probability change if there were only 6 lunches instead of 8?
2. What is the probability that all 5 friends pick the same lunch?
3. How would the solution change if there were 6 friends instead of 5?
4. What is the probability that exactly two friends will choose the same lunch?
5. How can this problem be approached using the inclusion-exclusion principle?
6. If each lunch costs $10, what is the expected total amount spent by all friends?
7. How does the probability change if one friend always orders the same lunch?
8. How would this problem differ if the friends choose lunches without replacement?

**Tip:** The complementary probability approach is often useful when calculating the probability of events with constraints

Explore the probability that two or more friends will order the same lunch when five friends choose from eight different lunch options. Use the complementary probability approach to calculate the likelihood of overlapping lunch choices.

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