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A book can be classified as either non-fiction or fiction. Suppose that \( 86 \% \) of books are classified as fiction. (a) Two books are chosen at random. What is the probability that both books are fiction? (b) Five books are chosen at random. What is the probability that all five books are fiction? (c) What is the probability that at least one of five randomly selected books is non-fiction? Would it be unusual that at least one of five randomly selected books is non-fiction? (a) Two books are chosen at random. What is the probability that both books are fiction? The probability is 7396 . (Round to four decimal places as needed.) (b) Five books are chosen at random. What is the probability that all five books are fiction? The probability is .4182 . (Round to four decimal places as needed.) (c) What is the probability that at least one of five randomly selected books is non-fiction? The probability is .5818 . (Round to four decimal places as needed.) Would it be unusual that at least one of five randomly selected books is non-fiction? It would not be unusual that at least one of five randomly selected books is non-fiction.

Ask by Romero Bob. in the United States
Feb 25,2025

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(a) The probability that both books are fiction is 0.7396. (b) The probability that all five books are fiction is 0.4182. (c) The probability that at least one of five randomly selected books is non-fiction is 0.5818. It would not be unusual for at least one of five randomly selected books to be non-fiction.

Solution

(a) The probability that both books are fiction is given by the formula: \[ P(\text{both fiction}) = P(\text{first book is fiction}) \times P(\text{second book is fiction}) \] Given that \( 86\% \) of books are classified as fiction, the probability of a book being fiction is \( 0.86 \). Therefore, the probability that both books are fiction is: \[ P(\text{both fiction}) = 0.86 \times 0.86 = 0.7396 \] (b) The probability that all five books are fiction is given by the formula: \[ P(\text{all fiction}) = P(\text{first book is fiction}) \times P(\text{second book is fiction}) \times P(\text{third book is fiction}) \times P(\text{fourth book is fiction}) \times P(\text{fifth book is fiction}) \] Using the same probability of a book being fiction as in part (a), the probability that all five books are fiction is: \[ P(\text{all fiction}) = 0.86 \times 0.86 \times 0.86 \times 0.86 \times 0.86 = 0.4182 \] (c) The probability that at least one of five randomly selected books is non-fiction is given by the formula: \[ P(\text{at least one non-fiction}) = 1 - P(\text{all fiction}) \] Using the probability of all five books being fiction from part (b), the probability that at least one of five randomly selected books is non-fiction is: \[ P(\text{at least one non-fiction}) = 1 - 0.4182 = 0.5818 \] It would not be unusual that at least one of five randomly selected books is non-fiction.

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The Deep Dive

Let’s take a lively journey into the world of probabilities! Imagine you go to a library where 86% of the books are fictional stories that transport you to magical lands, while 14% are non-fiction that enrich your understanding of the real world. To calculate the chance of picking books within these categories, we use the success probability of 0.86. In (a), picking two fiction books involves the equation \( P(A \cap B) = P(A) \times P(B) \), leading to \( 0.86 \times 0.86 = 0.7396 \). Likewise, for five fiction books in (b), the same logic applies, resulting in \( 0.86^5 \approx 0.4182 \). Now onto your question in (c)! To find out the chance of at least one non-fiction book among five selections, we flip the coin and focus on the complement of all being fiction: \( 1 - P(all fiction) = 1 - 0.4182 = 0.5818 \). With such a robust probability, encountering a non-fiction gem isn't just possible—it’s practically expected! So no, it would definitely not be unusual to find at least one lighthouse in your sea of novels!

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