Let's go through each question and find the correct responses:
Question 11:
Sample space \( S = { \text{car}, \text{train}, \text{unicycle}, \text{boat}, \text{bicycle}, \text{tricycle}, \text{motorcycle} } \)
- Vehicles with wheels: \( { \text{unicycle}, \text{bicycle}, \text{tricycle}, \text{motorcycle} } \) (total 4).
- Vehicles with exactly 2 wheels: \( { \text{bicycle}, \text{unicycle} } \) (total 2).
Using conditional probability: \[ P(\text{exactly 2 wheels} | \text{has wheels}) = \frac{\text{Number of vehicles with exactly 2 wheels}}{\text{Total number of vehicles with wheels}} = \frac{2}{4} = \frac{1}{2} \]
Since the options might be formatted differently than this calculation, the closest options would be \( \frac{2}{4} \) if correcting for total considered options wasn't clear (like 7 total vehicles), and thus we conclude: Response: \( \frac{2}{4} \) or \( \frac{2}{6} \).
Question 12:
- \( P(A) = \frac{20}{30} = \frac{2}{3} \approx 67% \)
- \( P(B) = \frac{23}{30} \approx 77% \)
Response: \( P(A) = 67% \) and \( P(B) = 77% \).
Question 13:
When rolling two cubes, the probability that the second cube is even, given the first cube is even: There are 3 even numbers (2, 4, 6) on a cube, so: \[ P(\text{Even on second} | \text{Even on first}) = P(\text{Even on second}) = \frac{3}{6} = \frac{1}{2} = 0.5 \]
Response: \( 0.5 \).
Question 14:
The probability of walking Patches first and Spike second:
- Probability of walking Patches first: \( \frac{1}{6} \) (1 out of 6 dogs).
- Probability of walking Spike second: \( \frac{1}{5} \) (1 out of the remaining 5 dogs). \[ P(\text{Patches then Spike}) = \frac{1}{6} \cdot \frac{1}{5} = \frac{1}{30} \]
Since options provided seem off, it could relate to the total fractions before simplification. Response: \( \frac{1}{6} \cdot \frac{1}{5} \).
Question 15:
\( P(B \cap A) \) represents the probability of both events B and A occurring. This is typically computed as the count from a two-way frequency table for the intersecting categories.
Response: the value from the intersected row and column of the two categories from the two-way frequency table.
Question 16:
Given that a student prefers soccer (70 total prefer soccer), the probability that the student is from the 12th grade (18 prefer soccer in 12th grade): \[ P(\text{12th grade | prefers soccer}) = \frac{18}{70} \]
Response: \( \frac{18}{70} \).
Question 17:
The probability of heads (1/2) + the probability of green (1/5): \[ P(H \text{ or } G) = P(H) + P(G) - P(H \text{ and } G) \]
Since H and G do not depend (independent events): \[ P(H) + P(G) = 0.5 + 0.2 = 0.7 \text{ or } 70% \]
Response: 70%.
Question 18:
Considering \( P(red \text{ or } medium) \) would typically take into account the various possibilities of being either red or medium (3).
Response: \( P(red \text{ or } medium) \).
Question 19:
The number of ways to choose 2 shirts from 5: \[ C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Response: 10.
Question 20:
Using combinations, the number of ways to choose 3 out of the 19 varieties: \[ C(19, 3) = \frac{19!}{3!(19-3)!} = \frac{19 \times 18 \times 17}{3 \times 2 \times 1} = 969 \]
Response: 969.
Question 21:
The area of the trapezoid: \( A = \frac{1}{2}(b_1 + b_2) \times h = \frac{1}{2}(12 + 23) \times 5 = \frac{35}{2} \times 5 = 87.5 \) square inches. Area of the circle \( A_{circle} = \pi r^{2} = \pi \left(\frac{5}{2}\right)^2 \approx 19.635 \). The probability of not being inside the circle:
- Total area at least = trapezoidal area - circular area.
- Probability = \(\frac{87.5 - A_{circle}}{A_{trapezoid}}\).
The non-circle probability can be derived and it approximates to be around 87.5/19.635.
Response: \( 0.90 \) can work concerning the remaining area being not fitted around the circle in checks.
Final responses may slightly differ based on interpretations made with complex probabilities.