Drag each tile to the correct box.

The table shows data on an airline's adherence to scheduled departure times over the weekend at an airport for 300 randomly selected flights.

\begin{tabular}{|c|c|c|c|}
\cline { 2 - 4 } \multicolumn{1}{c|}{} & On Time & Delayed & Total \\
\hline Domestic Flights & 142 & 68 & 210 \\
\hline International Flights & 56 & 34 & 90 \\
\hline Total & 198 & 102 & 300 \\
\hline
\end{tabular}

Order the probabilities of the given events from least to greatest.

- [tex]$P$[/tex] (on time and domestic)
- [tex]$P$[/tex] (delayed | domestic)
- [tex]$P$[/tex] (international | on time)

[tex]$\square$[/tex] [tex]$\ \textless \ $[/tex] [tex]$\square$[/tex] [tex]$\ \textless \ $[/tex] [tex]$\square$[/tex]



Answer :

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]