\begin{tabular}{l} Permutations and Probability \\ Determine the following probabilities. Enter your answers as percents rounded to four decimal \\ places. \\ The student council has 9 members where 5 of the members are Seniors. They need to choose a \\ President, Vice President, Secretary and Treasurer. \\ If the members are randomly assigned to the positions, what is the probability that the President is a \\ Senior? \\ \( \begin{array}{l}55.56 \\ \begin{array}{l}\text { Jacy has } 13 \text { books and he wants to read } 4 \text { over the summer. He has } 9 \text { fiction books and } 4 \text { nonfiction } \\ \text { books. } \\ \text { If Jacy randomly selects the } 4 \text { books, what is the probability that the first } 3 \text { books are fiction and the } \\ \text { next } 1 \text { books are nonfiction? } \\ \text { \% }\end{array}\end{array} \begin{array}{l}\text { \% }\end{array} \) \\ \hline\end{tabular}
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To find the probability that the President is a Senior, we first note that there are 9 members total and 5 of them are Seniors. The probability is calculated by taking the number of favorable outcomes over the total outcomes, resulting in \( P(\text{Senior President}) = \frac{5}{9} = 0.5556 \) or \( 55.56\% \). Now, for Jacy’s book selection, when he selects 4 books, we want the probability that the first 3 are fiction and the 4th is nonfiction. The calculation involves sequential selection without replacement. The probability for the first book being fiction is \( \frac{9}{13} \), the second \( \frac{8}{12} \), the third \( \frac{7}{11} \), and the last book being nonfiction is \( \frac{4}{10} \). Multiplying these gives us \( P = \frac{9}{13} \cdot \frac{8}{12} \cdot \frac{7}{11} \cdot \frac{4}{10} = 0.1294 \) or \( 12.94\% \). So, Jacy has approximately a 12.94% chance of pulling 3 fiction books followed by 1 nonfiction book!