Question

the two - way frequency table shows the current inventory of hardwood that a lumberyard carries. the hardwood board dimensions are given (e.g., (1\times2) means \1 foot by 2 feet\) along with their types.

suppose a board is chosen at random from the inventory, and it is a maple board. does this choice affect the probability of choosing a (1\times4) board from the inventory?

(p(\text{maple})cdot p(1\times2)=square)

the events \choosing a maple board\ and \choosing a (1\times2) board\ are independent not independent. a maple board has the same does not have the same probability of being chosen from all the boards as it has of being chosen from just the (1\times2) boards.

Explanation:

Step1: Calculate \(P(\text{maple})\)

The formula for probability is \(P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\).
The total number of boards is \(n = 210\), and the number of maple boards is \(n_{\text{maple}}=70\). So \(P(\text{maple})=\frac{70}{210}=\frac{1}{3}\).

Step2: Calculate \(P(1\times2)\)

The number of \(1\times2\) boards is \(n_{1\times2}=42\). So \(P(1\times2)=\frac{42}{210}=\frac{1}{5}\).

Step3: Calculate \(P(\text{maple})\cdot P(1\times2)\)

Multiply the two probabilities: \(\frac{1}{3}\times\frac{1}{5}=\frac{1}{15}\approx0.067\).

Step4: Check independence

The formula for independence is \(P(A\cap B)=P(A)\cdot P(B)\).
The number of maple \(1\times2\) boards is \(n_{\text{maple}\cap1\times2} = 14\). So \(P(\text{maple}\cap1\times2)=\frac{14}{210}=\frac{1}{15}\).
Since \(P(\text{maple}\cap1\times2)=P(\text{maple})\cdot P(1\times2)\), the events are independent.

For the last part, \(P(\text{maple})=\frac{70}{210}=\frac{1}{3}\), and \(P(\text{maple}\mid1\times2)=\frac{14}{42}=\frac{1}{3}\). So a maple board has the same probability of being chosen from all the boards as it has of being chosen from just the \(1\times2\) boards.

Answer:

\(P(\text{maple})\cdot P(1\times2)=\frac{1}{15}\approx0.067\)
The events “choosing a maple board” and “choosing a \(1\times2\) board” are independent. A maple board has the same probability of being chosen from all the boards as it has of being chosen from just the \(1\times2\) boards.
If a board is chosen at random and it is a maple board, this choice does not affect the probability of choosing a \(1\times4\) board from the inventory (because the two - events (choosing a maple board and choosing a \(1\times4\) board) are independent. The probability of choosing a \(1\times4\) board is \(P(1\times4)=\frac{54}{210}=\frac{9}{35}\), and \(P(1\times4\mid\text{maple})=\frac{18}{70}=\frac{9}{35}\))