analyzing a claim\nthe two - way table shows the distribution of book style to genre. miguel claims that…

analyzing a claim\nthe two - way table shows the distribution of book style to genre. miguel claims that given that the book is paperback (pb) does not affect the outcome that the book is nonfiction (nf). is miguels claim correct?\nfiction nonfiction total\npaperback 20 60 80\nhardcover 10 30 40\ntotal 30 90 120\nyes, the two events are independent because p(nf|pb)=p(nf).\nyes, the two events are independent because p(pb|nf)=p(nf).\nno, the two events are not independent because p(pb|nf)≠p(pb).\nno, the two events are not independent because p(pb|nf)≠p(nf).

analyzing a claim\nthe two - way table shows the distribution of book style to genre. miguel claims that given that the book is paperback (pb) does not affect the outcome that the book is nonfiction (nf). is miguels claim correct?\nfiction nonfiction total\npaperback 20 60 80\nhardcover 10 30 40\ntotal 30 90 120\nyes, the two events are independent because p(nf|pb)=p(nf).\nyes, the two events are independent because p(pb|nf)=p(nf).\nno, the two events are not independent because p(pb|nf)≠p(pb).\nno, the two events are not independent because p(pb|nf)≠p(nf).

Answer

Explanation:

Step1: Recall the definition of independent events

Two events $A$ and $B$ are independent if $P(A|B)=P(A)$. Here, let event $A$ be the book is non - fiction ($NF$) and event $B$ be the book is paperback ($PB$). We need to check if $P(NF|PB)=P(NF)$. The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In terms of the table values, $P(NF|PB)=\frac{n(NF\cap PB)}{n(PB)}$, where $n(NF\cap PB)$ is the number of non - fiction paperbacks and $n(PB)$ is the number of paperbacks. Also, $P(NF)=\frac{n(NF)}{n(Total)}$.

Step2: Calculate $P(NF|PB)$

$n(NF\cap PB) = 60$, $n(PB)=80$. So $P(NF|PB)=\frac{60}{80}=\frac{3}{4}$.

Step3: Calculate $P(NF)$

$n(NF)=90$, $n(Total) = 120$. So $P(NF)=\frac{90}{120}=\frac{3}{4}$. Since $P(NF|PB)=P(NF)$, the two events are independent.

Answer:

Yes, the two events are independent because $P(NF|PB)=P(NF)$.