Step 1: Interpret the problem.
We are asked to compute total daily home-based trips for:
- Households with size two and above (i.e., 2 persons or 3+ persons).
- Households with one or more vehicles (i.e., categories "One (1)" and "Two \& above (2+)").
We use Table I (trip rates per household) and Table II (number of households).
Step 2: Extract relevant values.
From Table I – Trip rate of households:
\[
\begin{array}{|c|c|c|}
\hline
\text{Vehicles/household} & \text{Two (2)} & \text{Three \& above (3+)}
\hline
\text{One (1)} & 2.5 & 5
\text{Two \& above (2+)} & 3 & 6
\hline
\end{array}
\]
From Table II – Household composition:
\[
\begin{array}{|c|c|c|}
\hline
\text{Vehicles/household} & \text{Two (2)} & \text{Three \& above (3+)}
\hline
\text{One (1)} & 300 & 200
\text{Two \& above (2+)} & 100 & 50
\hline
\end{array}
\]
Step 3: Compute trips = households × trip rate. \[ \text{Case 1: One vehicle, Two persons} = 300 \times 2.5 = 750 \] \[ \text{Case 2: One vehicle, Three+ persons} = 200 \times 5 = 1000 \] \[ \text{Case 3: Two+ vehicles, Two persons} = 100 \times 3 = 300 \] \[ \text{Case 4: Two+ vehicles, Three+ persons} = 50 \times 6 = 300 \]
Step 4: Add all contributions. \[ \text{Total Trips} = 750 + 1000 + 300 + 300 = 2350 \] Wait — let us double-check carefully.
Step 5: Careful rechecking with Table II. - One vehicle, Two persons → \(300 \times 2.5 = 750\) - One vehicle, 3+ persons → \(200 \times 5 = 1000\) - Two+ vehicles, Two persons → \(100 \times 3 = 300\) - Two+ vehicles, 3+ persons → \(50 \times 6 = 300\) Sum = \(2350\).
Final Answer: \[ \boxed{2350} \]
P and Q play chess frequently against each other. Of these matches, P has won 80% of the matches, drawn 15% of the matches, and lost 5% of the matches.
If they play 3 more matches, what is the probability of P winning exactly 2 of these 3 matches?