Answer :
Certainly! Let's walk through the problem step-by-step to find the probability that the group of stores will include a restaurant or a hardware store, or both.
First, we need to determine the total number of possible combinations of stores. Each store has a specific set of choices:
- Store 1 has 2 options: Pet Store or Restaurant.
- Store 2 has 3 options: Pet Store, Restaurant, or Gym.
- Store 3 has 2 options: Hardware Store or Pizza Shop.
The total number of combinations can be found by multiplying the number of options for each store:
[tex]\[ 2 \text{ options for Store 1} \times 3 \text{ options for Store 2} \times 2 \text{ options for Store 3} = 12 \][/tex]
Now, let's consider the combinations that include a restaurant, a hardware store, or both.
### Combinations including a restaurant
#### Restaurant in Store 1:
Here, we keep Store 1 fixed as a restaurant and vary the other stores:
[tex]\[ 1 \text{ (option for Store 1 being fixed as Restaurant)} \times 3 \text{ (options for Store 2)} \times 2 \text{ (options for Store 3)} = 6 \][/tex]
#### Restaurant in Store 2:
Here, we keep Store 2 fixed as a restaurant and vary the other stores:
[tex]\[ 2 \text{ (options for Store 1)} \times 1 \text{ (option for Store 2 being fixed as Restaurant)} \times 2 \text{ (options for Store 3)} = 4 \][/tex]
Summing these, the total number of combinations including a restaurant:
[tex]\[ 6 + 4 = 10 \][/tex]
### Combinations including a hardware store
#### Hardware Store in Store 3:
Here, we keep Store 3 fixed as a hardware store and vary the other stores:
[tex]\[ 2 \text{ (options for Store 1)} \times 3 \text{ (options for Store 2)} \times 1 \text{ (option for Store 3 being fixed as Hardware Store)} = 6 \][/tex]
### Adjusting for over-counting of combinations that include both a restaurant and a hardware store:
We need to subtract the combinations that include both a restaurant and a hardware store from our count, as they've been counted twice (once in the restaurant count and once in the hardware store count).
#### Both Restaurant and Hardware Store:
- Restaurant in Store 1 and Hardware Store in Store 3:
[tex]\[ 1 \text{ (option for Store 1 fixed as Restaurant)} \times 3 \text{ (Store 2 options)} \times 1 \text{ (option for Store 3 fixed as Hardware Store)} = 3 \][/tex]
- Restaurant in Store 2 and Hardware Store in Store 3:
[tex]\[ 2 \text{ (Store 1 options)} \times 1 \text{ (option for Store 2 fixed as Restaurant)} \times 1 \text{ (option for Store 3 fixed as Hardware Store)} = 2 \][/tex]
Summing these, the total number of combinations including both a restaurant and a hardware store:
[tex]\[ 3 + 2 = 5 \][/tex]
### Combinations including a restaurant or a hardware store or both:
To find the total number of favorable combinations, we add the total combinations including a restaurant and the total combinations including a hardware store, then subtract the over-counted combinations that include both:
[tex]\[ 10 \text{ (including a restaurant)} + 6 \text{ (including a hardware store)} - 5 \text{ (including both)} = 11 \][/tex]
### Probability
Finally, the probability that the set of stores includes a restaurant or a hardware store, or both, is the number of favorable combinations over the total number of combinations:
[tex]\[ \frac{11}{12} \][/tex]
Therefore, the probability that the group of stores will have a restaurant or a hardware store, or both, is:
[tex]\[ \frac{11}{12} \][/tex]
First, we need to determine the total number of possible combinations of stores. Each store has a specific set of choices:
- Store 1 has 2 options: Pet Store or Restaurant.
- Store 2 has 3 options: Pet Store, Restaurant, or Gym.
- Store 3 has 2 options: Hardware Store or Pizza Shop.
The total number of combinations can be found by multiplying the number of options for each store:
[tex]\[ 2 \text{ options for Store 1} \times 3 \text{ options for Store 2} \times 2 \text{ options for Store 3} = 12 \][/tex]
Now, let's consider the combinations that include a restaurant, a hardware store, or both.
### Combinations including a restaurant
#### Restaurant in Store 1:
Here, we keep Store 1 fixed as a restaurant and vary the other stores:
[tex]\[ 1 \text{ (option for Store 1 being fixed as Restaurant)} \times 3 \text{ (options for Store 2)} \times 2 \text{ (options for Store 3)} = 6 \][/tex]
#### Restaurant in Store 2:
Here, we keep Store 2 fixed as a restaurant and vary the other stores:
[tex]\[ 2 \text{ (options for Store 1)} \times 1 \text{ (option for Store 2 being fixed as Restaurant)} \times 2 \text{ (options for Store 3)} = 4 \][/tex]
Summing these, the total number of combinations including a restaurant:
[tex]\[ 6 + 4 = 10 \][/tex]
### Combinations including a hardware store
#### Hardware Store in Store 3:
Here, we keep Store 3 fixed as a hardware store and vary the other stores:
[tex]\[ 2 \text{ (options for Store 1)} \times 3 \text{ (options for Store 2)} \times 1 \text{ (option for Store 3 being fixed as Hardware Store)} = 6 \][/tex]
### Adjusting for over-counting of combinations that include both a restaurant and a hardware store:
We need to subtract the combinations that include both a restaurant and a hardware store from our count, as they've been counted twice (once in the restaurant count and once in the hardware store count).
#### Both Restaurant and Hardware Store:
- Restaurant in Store 1 and Hardware Store in Store 3:
[tex]\[ 1 \text{ (option for Store 1 fixed as Restaurant)} \times 3 \text{ (Store 2 options)} \times 1 \text{ (option for Store 3 fixed as Hardware Store)} = 3 \][/tex]
- Restaurant in Store 2 and Hardware Store in Store 3:
[tex]\[ 2 \text{ (Store 1 options)} \times 1 \text{ (option for Store 2 fixed as Restaurant)} \times 1 \text{ (option for Store 3 fixed as Hardware Store)} = 2 \][/tex]
Summing these, the total number of combinations including both a restaurant and a hardware store:
[tex]\[ 3 + 2 = 5 \][/tex]
### Combinations including a restaurant or a hardware store or both:
To find the total number of favorable combinations, we add the total combinations including a restaurant and the total combinations including a hardware store, then subtract the over-counted combinations that include both:
[tex]\[ 10 \text{ (including a restaurant)} + 6 \text{ (including a hardware store)} - 5 \text{ (including both)} = 11 \][/tex]
### Probability
Finally, the probability that the set of stores includes a restaurant or a hardware store, or both, is the number of favorable combinations over the total number of combinations:
[tex]\[ \frac{11}{12} \][/tex]
Therefore, the probability that the group of stores will have a restaurant or a hardware store, or both, is:
[tex]\[ \frac{11}{12} \][/tex]
