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To solve this problem, we'll use the law of total probability. The law of total probability states that if we have multiple mutually exclusive events that partition the sample space, we can find the probability of a union of these events by summing the probabilities of each event multiplied by the conditional probability of the interested event given each event. Let's break it down step-by-step.
1. Determine the probability of selecting each factory:
- Probability of selecting factory A ([tex]\( P(A) \)[/tex]) is [tex]\( 0.50 \)[/tex]
- Probability of selecting factory B ([tex]\( P(B) \)[/tex]) is [tex]\( 0.25 \)[/tex]
- Probability of selecting factory C ([tex]\( P(C) \)[/tex]) is [tex]\( 0.25 \)[/tex]
2. Determine the conditional probability of producing a passenger airplane at each factory:
- Probability of a passenger airplane given it's from factory A ([tex]\( P(\text{Passenger} | A) \)[/tex]) is [tex]\( 0.70 \)[/tex]
- Probability of a passenger airplane given it's from factory B ([tex]\( P(\text{Passenger} | B) \)[/tex]) is [tex]\( 0.25 \)[/tex]
- Probability of a passenger airplane given it's from factory C ([tex]\( P(\text{Passenger} | C) \)[/tex]) is [tex]\( 0.25 \)[/tex]
3. Apply the law of total probability to find the overall probability that the airplane is a passenger airplane:
[tex]\[ P(\text{Passenger}) = P(A) \cdot P(\text{Passenger} | A) + P(B) \cdot P(\text{Passenger} | B) + P(C) \cdot P(\text{Passenger} | C) \][/tex]
4. Plug in the values to get the result:
[tex]\[ P(\text{Passenger}) = (0.50 \times 0.70) + (0.25 \times 0.25) + (0.25 \times 0.25) \][/tex]
[tex]\[ P(\text{Passenger}) = (0.50 \times 0.70) + (0.25 \times 0.25) + (0.25 \times 0.25) \][/tex]
[tex]\[ P(\text{Passenger}) = 0.35 + 0.0625 + 0.0625 \][/tex]
[tex]\[ P(\text{Passenger}) = 0.475 \][/tex]
Therefore, the probability that a randomly selected airplane is a passenger airplane is:
A. 0.475
So, the correct answer is 0.475.
1. Determine the probability of selecting each factory:
- Probability of selecting factory A ([tex]\( P(A) \)[/tex]) is [tex]\( 0.50 \)[/tex]
- Probability of selecting factory B ([tex]\( P(B) \)[/tex]) is [tex]\( 0.25 \)[/tex]
- Probability of selecting factory C ([tex]\( P(C) \)[/tex]) is [tex]\( 0.25 \)[/tex]
2. Determine the conditional probability of producing a passenger airplane at each factory:
- Probability of a passenger airplane given it's from factory A ([tex]\( P(\text{Passenger} | A) \)[/tex]) is [tex]\( 0.70 \)[/tex]
- Probability of a passenger airplane given it's from factory B ([tex]\( P(\text{Passenger} | B) \)[/tex]) is [tex]\( 0.25 \)[/tex]
- Probability of a passenger airplane given it's from factory C ([tex]\( P(\text{Passenger} | C) \)[/tex]) is [tex]\( 0.25 \)[/tex]
3. Apply the law of total probability to find the overall probability that the airplane is a passenger airplane:
[tex]\[ P(\text{Passenger}) = P(A) \cdot P(\text{Passenger} | A) + P(B) \cdot P(\text{Passenger} | B) + P(C) \cdot P(\text{Passenger} | C) \][/tex]
4. Plug in the values to get the result:
[tex]\[ P(\text{Passenger}) = (0.50 \times 0.70) + (0.25 \times 0.25) + (0.25 \times 0.25) \][/tex]
[tex]\[ P(\text{Passenger}) = (0.50 \times 0.70) + (0.25 \times 0.25) + (0.25 \times 0.25) \][/tex]
[tex]\[ P(\text{Passenger}) = 0.35 + 0.0625 + 0.0625 \][/tex]
[tex]\[ P(\text{Passenger}) = 0.475 \][/tex]
Therefore, the probability that a randomly selected airplane is a passenger airplane is:
A. 0.475
So, the correct answer is 0.475.
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