Question

A large company plans to develop a private subdivision. There are 6house models o choose from and for each model, a client may opt to have a garage or carport. Use A, B, C, D E, and F as labels for the house models. generate frequencies that can be used to approximate the probability of a client who will choose model B with a car garage.
Use result of 100 trials.
A-10
B- 16
C- 20
D- 14
E- 30
F- 10

Answer 1

To calculate the probability of a client choosing model B with a car garage, we first need to use the provided frequencies of house model selections to establish the number of times model B is chosen.

### House Model Frequencies (Out of 100 trials):
- A: 10
- B: 16
- C: 20
- D: 14
- E: 30
- F: 10

### Total:
The total number of trials is 100, as you mentioned.

To find the frequency of clients choosing model B, we can use the frequency given (16) and note that each model can have either a garage or a carport. However, there's no specific frequency provided for the garage option; therefore, we'll assume that each option is equally likely (i.e., half the clients chose a garage and half chose a carport).

From this approach:
- Total clients choosing model B = 16.
- Assuming equal preference for garage and carport:
- Clients choosing model B with a garage = 16 / 2 = 8.
- Clients choosing model B with a carport = 16 / 2 = 8.

### Probability Calculation:
The probability of a client choosing model B with a garage can be calculated as follows:

\[
P(B \text{ with garage}) = \frac{\text{Number of clients choosing B with garage}}{\text{Total number of trials}} = \frac{8}{100} = 0.08
\]

### Conclusion:
The approximate probability of a client choosing model B with a car garage is **0.08** or **8%**.