Answer :
To determine if the events "being a nonfiction book (NF)" and "being a paperback book (PB)" are independent, we need to analyze the probabilities involved and compare conditional probabilities with marginal probabilities.
Let's define the events:
- [tex]\( P(NF) \)[/tex]: Probability that a book is nonfiction.
- [tex]\( P(PB) \)[/tex]: Probability that a book is paperback.
- [tex]\( P(PB \mid NF) \)[/tex]: Probability that a book is paperback given that it is nonfiction.
First, let's calculate the necessary probabilities based on the provided two-way table:
[tex]\[ \begin{tabular}{|c|c|c|c|} \cline { 2 - 4 } & Fiction & Nonfiction & Total \\ \hline Paperback & 20 & 60 & 80 \\ \hline Hardcover & 10 & 30 & 40 \\ \hline Total & 30 & 90 & 120 \\ \hline \end{tabular} \][/tex]
1. Calculate [tex]\( P(NF) \)[/tex]:
[tex]\[ P(NF) = \frac{\text{Number of nonfiction books}}{\text{Total number of books}} = \frac{90}{120} = 0.75 \][/tex]
2. Calculate [tex]\( P(PB) \)[/tex]:
[tex]\[ P(PB) = \frac{\text{Number of paperback books}}{\text{Total number of books}} = \frac{80}{120} = \frac{2}{3} \approx 0.67 \][/tex]
3. Calculate [tex]\( P(PB \mid NF) \)[/tex]:
[tex]\[ P(PB \mid NF) = \frac{\text{Number of paperback nonfiction books}}{\text{Total number of nonfiction books}} = \frac{60}{90} = \frac{2}{3} \approx 0.67 \][/tex]
To check if the events are independent, we compare [tex]\( P(PB \mid NF) \)[/tex] and [tex]\( P(PB) \)[/tex]. If [tex]\( P(PB \mid NF) = P(PB) \)[/tex], then the events are independent.
Given the calculated probabilities:
[tex]\[ P(PB \mid NF) = \frac{2}{3} \approx 0.67 \][/tex]
[tex]\[ P(PB) = \frac{2}{3} \approx 0.67 \][/tex]
Since [tex]\( P(PB \mid NF) = P(PB) \)[/tex], the events "being a nonfiction book (NF)" and "being a paperback book (PB)" are indeed independent.
Hence, Miguel's claim is correct.
Therefore, the correct answer is:
Yes, the two events are independent because [tex]\( P(PB \mid NF) = P(PB) \)[/tex].
Let's define the events:
- [tex]\( P(NF) \)[/tex]: Probability that a book is nonfiction.
- [tex]\( P(PB) \)[/tex]: Probability that a book is paperback.
- [tex]\( P(PB \mid NF) \)[/tex]: Probability that a book is paperback given that it is nonfiction.
First, let's calculate the necessary probabilities based on the provided two-way table:
[tex]\[ \begin{tabular}{|c|c|c|c|} \cline { 2 - 4 } & Fiction & Nonfiction & Total \\ \hline Paperback & 20 & 60 & 80 \\ \hline Hardcover & 10 & 30 & 40 \\ \hline Total & 30 & 90 & 120 \\ \hline \end{tabular} \][/tex]
1. Calculate [tex]\( P(NF) \)[/tex]:
[tex]\[ P(NF) = \frac{\text{Number of nonfiction books}}{\text{Total number of books}} = \frac{90}{120} = 0.75 \][/tex]
2. Calculate [tex]\( P(PB) \)[/tex]:
[tex]\[ P(PB) = \frac{\text{Number of paperback books}}{\text{Total number of books}} = \frac{80}{120} = \frac{2}{3} \approx 0.67 \][/tex]
3. Calculate [tex]\( P(PB \mid NF) \)[/tex]:
[tex]\[ P(PB \mid NF) = \frac{\text{Number of paperback nonfiction books}}{\text{Total number of nonfiction books}} = \frac{60}{90} = \frac{2}{3} \approx 0.67 \][/tex]
To check if the events are independent, we compare [tex]\( P(PB \mid NF) \)[/tex] and [tex]\( P(PB) \)[/tex]. If [tex]\( P(PB \mid NF) = P(PB) \)[/tex], then the events are independent.
Given the calculated probabilities:
[tex]\[ P(PB \mid NF) = \frac{2}{3} \approx 0.67 \][/tex]
[tex]\[ P(PB) = \frac{2}{3} \approx 0.67 \][/tex]
Since [tex]\( P(PB \mid NF) = P(PB) \)[/tex], the events "being a nonfiction book (NF)" and "being a paperback book (PB)" are indeed independent.
Hence, Miguel's claim is correct.
Therefore, the correct answer is:
Yes, the two events are independent because [tex]\( P(PB \mid NF) = P(PB) \)[/tex].
