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Sagot :
Sure, let's break down the given data and the required probabilities step-by-step.
1. Data Interpretation from the table:
- On Time Domestic Flights: 142
- Delayed Domestic Flights: 68
- Total Domestic Flights: 210
- On Time International Flights: 56
- Total Flights: 300
2. Calculations:
1. Probability of a flight being on time and domestic, [tex]\( P(\text{on time and domestic}) \)[/tex]:
- The total number of flights is 300.
- The number of on time domestic flights is 142.
[tex]\[ P(\text{on time and domestic}) = \frac{142}{300} = 0.4733 \][/tex]
2. Probability of a delayed flight given that it is domestic, [tex]\( P(\text{delayed} \mid \text{domestic}) \)[/tex]:
- The total number of domestic flights is 210.
- The number of delayed domestic flights is 68.
[tex]\[ P(\text{delayed} \mid \text{domestic}) = \frac{68}{210} = 0.3238 \][/tex]
3. Probability of an international flight given that it is on time, [tex]\( P(\text{international} \mid \text{on time}) \)[/tex]:
- The total number of on time flights is 198 (142 domestic + 56 international).
- The number of on time international flights is 56.
[tex]\[ P(\text{international} \mid \text{on time}) = \frac{56}{198} = 0.2828 \][/tex]
3. Order the probabilities from least to greatest:
[tex]\[ 0.2828 < 0.3238 < 0.4733 \][/tex]
So, the correctly ordered probabilities are:
[tex]\[ P(\text{international} \mid \text{on time}) < P(\text{delayed} \mid \text{domestic}) < P(\text{on time and domestic}) \][/tex]
Therefore, the solution in terms of the given probabilities is:
[tex]\[ \text{box1} = P(\text{international} \mid \text{on time}), \quad \text{box2} = P(\text{delayed} \mid \text{domestic}), \quad \text{box3} = P(\text{on time and domestic}) \][/tex]
In order:
[tex]\[ \boxed{P(\text{international} \mid \text{on time})} < \boxed{P(\text{delayed} \mid \text{domestic})} < \boxed{P(\text{on time and domestic})} \][/tex]
1. Data Interpretation from the table:
- On Time Domestic Flights: 142
- Delayed Domestic Flights: 68
- Total Domestic Flights: 210
- On Time International Flights: 56
- Total Flights: 300
2. Calculations:
1. Probability of a flight being on time and domestic, [tex]\( P(\text{on time and domestic}) \)[/tex]:
- The total number of flights is 300.
- The number of on time domestic flights is 142.
[tex]\[ P(\text{on time and domestic}) = \frac{142}{300} = 0.4733 \][/tex]
2. Probability of a delayed flight given that it is domestic, [tex]\( P(\text{delayed} \mid \text{domestic}) \)[/tex]:
- The total number of domestic flights is 210.
- The number of delayed domestic flights is 68.
[tex]\[ P(\text{delayed} \mid \text{domestic}) = \frac{68}{210} = 0.3238 \][/tex]
3. Probability of an international flight given that it is on time, [tex]\( P(\text{international} \mid \text{on time}) \)[/tex]:
- The total number of on time flights is 198 (142 domestic + 56 international).
- The number of on time international flights is 56.
[tex]\[ P(\text{international} \mid \text{on time}) = \frac{56}{198} = 0.2828 \][/tex]
3. Order the probabilities from least to greatest:
[tex]\[ 0.2828 < 0.3238 < 0.4733 \][/tex]
So, the correctly ordered probabilities are:
[tex]\[ P(\text{international} \mid \text{on time}) < P(\text{delayed} \mid \text{domestic}) < P(\text{on time and domestic}) \][/tex]
Therefore, the solution in terms of the given probabilities is:
[tex]\[ \text{box1} = P(\text{international} \mid \text{on time}), \quad \text{box2} = P(\text{delayed} \mid \text{domestic}), \quad \text{box3} = P(\text{on time and domestic}) \][/tex]
In order:
[tex]\[ \boxed{P(\text{international} \mid \text{on time})} < \boxed{P(\text{delayed} \mid \text{domestic})} < \boxed{P(\text{on time and domestic})} \][/tex]
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