To determine if Miguel's claim is correct, we need to check whether the events "the book is a paperback (PB)" and "the book is a nonfiction (NF)" are independent.
Two events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent if and only if:
[tex]\[ P(A \mid B) = P(A) \][/tex]
In our context, this means that the probability of a book being nonfiction given that it is paperback ([tex]\(P(NF \mid PB)\)[/tex]) should be equal to the overall probability of a book being nonfiction ([tex]\(P(NF)\)[/tex]).
Let's calculate the probabilities step by step:
1. Calculate [tex]\( P(NF \mid PB) \)[/tex]:
- This is the probability that a book is nonfiction given that it is paperback.
- We use the numbers from the table:
- Nonfiction paperback books (PB & NF): 60
- Total paperback books (PB): 80
[tex]\[
P(NF \mid PB) = \frac{\text{Number of nonfiction paperback books}}{\text{Total number of paperback books}} = \frac{60}{80} = 0.75
\][/tex]
2. Calculate [tex]\( P(NF) \)[/tex]:
- This is the overall probability that a book is nonfiction.
- We use the numbers from the table:
- Total nonfiction books (NF): 90
- Total number of books: 120
[tex]\[
P(NF) = \frac{\text{Total number of nonfiction books}}{\text{Total number of books}} = \frac{90}{120} = 0.75
\][/tex]
3. Compare [tex]\( P(NF \mid PB) \)[/tex] and [tex]\( P(NF) \)[/tex]:
- We found that:
[tex]\[
P(NF \mid PB) = 0.75
\][/tex]
[tex]\[
P(NF) = 0.75
\][/tex]
Since [tex]\( P(NF \mid PB) = P(NF) \)[/tex], we conclude that the events "the book is a paperback (PB)" and "the book is a nonfiction (NF)" are indeed independent.
Therefore, Miguel's claim is correct. The correct statement is:
- Yes, the two events are independent because [tex]\(P(NF \mid PB) = P(NF)\)[/tex].
