To find the probability that an object has exactly two wheels given that the object has wheels, we first need to identify the vehicles in the sample space that have wheels:
- Car (4 wheels)
- Train (not a vehicle for this context but typically has many wheels and can be excluded)
- Unicycle (1 wheel)
- Boat (typically no wheels)
- Bicycle (2 wheels)
- Tricycle (3 wheels)
- Motorcycle (2 wheels)
The vehicles from the sample space that have wheels are:
- Unicycle (1 wheel)
- Bicycle (2 wheels)
- Tricycle (3 wheels)
- Motorcycle (2 wheels)
- Car (4 wheels)
Now, let's count how many of these vehicles have wheels:
- Unicycle: 1 wheel
- Bicycle: 2 wheels
- Tricycle: 3 wheels
- Motorcycle: 2 wheels
- Car: 4 wheels
So, the vehicles with wheels are:
- Unicycle,
- Bicycle,
- Tricycle,
- Motorcycle,
- and Car.
This gives us a total of 5 vehicles with wheels. Among these, the vehicles with exactly 2 wheels are:
- Bicycle
- Motorcycle
So, there are 2 vehicles with exactly 2 wheels.
Now we can compute the probability:
\[ P(\text{exactly 2 wheels} \mid \text{has wheels}) = \frac{\text{Number of vehicles with exactly 2 wheels}}{\text{Total number of vehicles with wheels}} = \frac{2}{5} \]
Now we look at the given response options:
- \( \frac{26}{6} \)
- \( \frac{17}{7} \)
- \( \frac{37}{7} \)
- \( \frac{27}{7} \)
None of these options match the probability of \( \frac{2}{5} \).
However, if we look at them in terms of a common denominator, we can approximately compare:
- \( \frac{26}{6} \approx 4.33 \)
- \( \frac{17}{7} \approx 2.43 \)
- \( \frac{37}{7} \approx 5.29 \)
- \( \frac{27}{7} \approx 3.86 \)
Since none match or seem to come close to \( 0.4 \) or \( \frac{2}{5} \), there may be an error in the framing of the question or the provided answers context.
In conclusion, the exact probability that an object has exactly two wheels given that it has wheels is \( \frac{2}{5} \) which is not present among the provided options.