Certainly! Let's solve the problem. We're given a situation where an insurance company surveyed 200 young adults about their first motor vehicle. We need to find the probability that a randomly selected participant had either a four-cylinder vehicle, a car, or both.
We can use some basic principles of probability to figure this out.
### Given Values:
- Total number of participants surveyed = 200
- Number of participants whose first motor vehicle had four cylinders ([tex]\( F \)[/tex]) = 76
- Number of participants whose first motor vehicle was a car ([tex]\( C \)[/tex]) = 145
- Number of participants whose first motor vehicle was a four-cylinder car ([tex]\( F \cap C \)[/tex]) = 104
### Step-by-Step Solution:
#### 1. Calculate the probabilities of each event individually:
- Probability that a participant's first motor vehicle had four cylinders ([tex]\( P(F) \)[/tex]):
[tex]\[
P(F) = \frac{\text{Number of participants with four-cylinder motor vehicle}}{\text{Total number of participants}} = \frac{76}{200} = 0.38
\][/tex]
- Probability that a participant's first motor vehicle was a car ([tex]\( P(C) \)[/tex]):
[tex]\[
P(C) = \frac{\text{Number of participants whose first motor vehicle was a car}}{\text{Total number of participants}} = \frac{145}{200} = 0.725
\][/tex]
#### 2. Calculate the probability that a participant's first motor vehicle was a four-cylinder car ([tex]\( P(F \cap C) \)[/tex]):
[tex]\[
P(F \cap C) = \frac{\text{Number of participants whose first motor vehicle was a four-cylinder car}}{\text{Total number of participants}} = \frac{104}{200} = 0.52
\][/tex]
#### 3. Use the formula for [tex]\( P(F \cup C) \)[/tex]:
The formula for the probability of the union of two events, also known as [tex]\( P(F \text{ or } C) \)[/tex], is given by:
[tex]\[
P(F \cup C) = P(F) + P(C) - P(F \cap C)
\][/tex]
Substituting the values we found:
[tex]\[
P(F \cup C) = 0.38 + 0.725 - 0.52 = 0.585
\][/tex]
### Conclusion:
The probability that a randomly selected participant's first motor vehicle had either four cylinders, was a car, or both, is:
[tex]\[
P(F \cup C) = 0.585
\][/tex]
Comparing this result with the given options:
- 0.59
- 0.68
- 0.71
- 0.80
None of the options exactly match 0.585, but 0.59 is the closest approximation. Thus, the answer is:
[tex]\[ \boxed{0.59} \][/tex]
