Question

frequency tables, probability, and independent events
the two - way frequency table shows the current inventory of hardwood that a lumberyard carries. the hardwood board dimensions are given (e.g., 1×2 means “1 foot by 2 feet”) along with their types.

	oak	maple	cherry	total
1×3	13	28	17	58
1×4	17	18	19	54
1×6	12	10	34	56
total	62	70	78	210

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 1×4 board from the inventory?
we need to know if the two events are independent or not. if the two events are independent, then:
(p(\text{maple and } 1×4)=p(\text{maple})cdot p(1×4)).
(p(\text{maple})=\frac{70}{210}), (p(1×4)=\frac{54}{210}), and (p(\text{maple and } 1×4)=\frac{18}{210})
(\frac{70}{210}cdot\frac{54}{210}=\frac{18}{210})?, so the two events, “choosing a maple board” and “choosing a 1×4 board” are independent events. knowing that a maple board was chosen at random from the inventory does not affect the probability of choosing a 1×4 board. this means that a 1×4 board has the same probability of being chosen from all the boards, (\frac{54}{210}), as it has of being chosen from just the maple boards, (\frac{18}{70}). also, a maple board has the same probability of being chosen from all the boards, (\frac{70}{210}), as it has of being chosen from just the 1×4 boards, (\frac{18}{54}).

study the worked example and then answer each question.
suppose a board is chosen at random from the inventory, and it is a 1×2 board. does this choice affect the probability of choosing a maple board from the inventory?
(p(\text{maple})=) (square)
(p(1×2)=) (square)
(p(\text{maple and } 1×2)=) (square)
(p(\text{maple})cdot p(1×2)=) (square)
the events “choosing a maple board” and “choosing a 1×2 board” are (\bigcirc) independent (\bigcirc) not independent
a maple board (\bigcirc) has the same (\bigcirc) does not have the same probability of being chosen from all the boards as it has of being chosen from just the 1×2 boards.

Explanation:

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

The total number of boards is 210, and the number of maple boards is 70. So \( P(\text{maple})=\frac{70}{210}=\frac{1}{3} \).

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

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

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

The number of maple and \( 1\times2 \) boards is 14. So \( P(\text{maple and } 1\times2)=\frac{14}{210}=\frac{1}{15} \).

Step4: Calculate \( P(\text{maple})\cdot P(1\times2) \)

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

Step5: Compare \( P(\text{maple and } 1\times2) \) and \( P(\text{maple})\cdot P(1\times2) \)

Since \( \frac{1}{15}=\frac{1}{15} \), the events are independent. Also, check the conditional probability: \( P(\text{maple} | 1\times2)=\frac{14}{42}=\frac{1}{3} \), which is equal to \( P(\text{maple})=\frac{1}{3} \), so the probability of choosing a maple board is the same whether we know it's a \( 1\times2 \) board or not.

Answer:

• \( P(\text{maple})=\frac{70}{210}=\frac{1}{3} \)
• \( P(1\times2)=\frac{42}{210}=\frac{1}{5} \)
• \( P(\text{maple and } 1\times2)=\frac{14}{210}=\frac{1}{15} \)
• \( P(\text{maple})\cdot P(1\times2)=\frac{1}{3}\times\frac{1}{5}=\frac{1}{15} \)
• The events "choosing a maple board" and "choosing a \( 1\times2 \) board" are independent, and 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.