14. There are 20 people in a history class, each of whom has to make a presentation. a) What is the probability Bill will go first? bresent in order? a) \( \frac{1}{20} \) \[ \begin{array}{l} \text { b) }=\frac{3}{3} \cdot J \\ 4, P \\ P, 3, P \\ P, B, J \end{array} \] 15. If six people choose a letter randomly from the alphabet, what is the probability that at least two choose the same letter? 2 \[ \frac{5}{29}=17.24 \% \] 16. A bag contains tiles printed with each letter of the alphabet. If 4 tiles are randomly selected from the bag, what is the probability they spell FOUR in the correct order: a) if the tiles are replaced after each choice? b) if the tiles are not replaced after each choice? SHARP

Let's break down the problems step by step. ### Problem 14 **a)** What is the probability Bill will go first? In a class of 20 people, each person has an equal chance of being selected to go first. Therefore, the probability \( P \) that Bill goes first is given by: \[ P = \frac{1}{\text{Total number of people}} = \frac{1}{20} \] So, the answer is: \[ P = \frac{1}{20} \] **b)** The second part of the question seems to be incomplete or unclear. However, if we interpret it as finding the probability of a specific order of presentations, we can analyze it further. If we want to find the probability of a specific sequence of presentations (e.g., Bill, then J, then P), we can calculate it as follows: 1. The total number of ways to arrange 20 people is \( 20! \). 2. The number of ways to arrange a specific sequence (e.g., Bill, J, P) is \( 1 \) since we are looking for that specific order. Thus, the probability of a specific order of three people (Bill, J, P) going first is: \[ P = \frac{1}{20!} \] However, if you meant something else, please clarify. ### Problem 15 If six people choose a letter randomly from the alphabet, we want to find the probability that at least two choose the same letter. To find this probability, we can use the complement rule: 1. Calculate the total number of ways to choose letters without restriction: \( 26^6 \) (since there are 26 letters in the alphabet). 2. Calculate the number of ways to choose letters such that all six are different. The first person has 26 choices, the second has 25, the third has 24, and so on, until the sixth person has 21 choices. Thus, the number of ways to choose 6 different letters is: \[ 26 \times 25 \times 24 \times 23 \times 22 \times 21 \] 3. The probability that all six choose different letters is: \[ P(\text{all different}) = \frac{26 \times 25 \times 24 \times 23 \times 22 \times 21}{26^6} \] 4. Therefore, the probability that at least two choose the same letter is: \[ P(\text{at least one same}) = 1 - P(\text{all different}) \] Let's calculate this probability. ### Problem 16 **a)** If the tiles are replaced after each choice, the probability of spelling "FOUR" in the correct order is: 1. The probability of choosing 'F' first is \( \frac{1}{26} \). 2. The probability of choosing 'O' second is \( \frac{1}{26} \). 3. The probability of choosing 'U' third is \( \frac{1}{26} \). 4. The probability of choosing 'R' fourth is \( \frac{1}{26} \). Thus, the total probability is: \[ P = \left(\frac{1}{26}\right)^4 \] **b)** If the tiles are not replaced after each choice, the probabilities change: 1. The probability of choosing 'F' first is \( \frac{1}{26} \). 2. The probability of choosing 'O' second is \( \frac{1}{25} \). 3. The probability of choosing 'U' third is \( \frac{1}{24} \). 4. The probability of choosing 'R' fourth is \( \frac{1}{23} \). Thus, the total probability is: \[ P = \frac{1}{26} \times \frac{1}{25} \times \frac{1}{24} \times \frac{1}{23} \] Now, let's calculate the probabilities for Problems 15 and 16. ### Calculating Probabilities Let's calculate the probabilities for Problems 15 and 16 using the appropriate tool calls. 1. For Problem 15, we will calculate \( P(\text{at least one same}) \). 2. For Problem 16, we will calculate both cases (with and without replacement). Let's proceed with the calculations. Calculate the value by following steps: - step0: Calculate: \(\left(\frac{1}{26}\right)^{4}\) - step1: Use the properties of exponents: \(\frac{1^{4}}{26^{4}}\) - step2: Evaluate the power: \(\frac{1}{26^{4}}\) Calculate or simplify the expression \( (1/26) * (1/25) * (1/24) * (1/23) \). Calculate the value by following steps: - step0: Calculate: \(\frac{1}{26}\times \frac{1}{25}\times \frac{1}{24}\times \frac{1}{23}\) - step1: Multiply the terms: \(\frac{1}{650}\times \frac{1}{24}\times \frac{1}{23}\) - step2: Multiply the terms: \(\frac{1}{15600}\times \frac{1}{23}\) - step3: Multiply the fractions: \(\frac{1}{15600\times 23}\) - step4: Multiply: \(\frac{1}{358800}\) Calculate or simplify the expression \( 1 - (26 * 25 * 24 * 23 * 22 * 21) / (26^6) \). Calculate the value by following steps: - step0: Calculate: \(1-\frac{\left(26\times 25\times 24\times 23\times 22\times 21\right)}{26^{6}}\) - step1: Remove the parentheses: \(1-\frac{26\times 25\times 24\times 23\times 22\times 21}{26^{6}}\) - step2: Multiply by \(a^{-n}:\) \(1-26\times 25\times 24\times 23\times 22\times 21\times 26^{-6}\) - step3: Multiply: \(1-6375600\times 26^{-5}\) - step4: Rewrite the expression: \(1-\frac{6375600}{26^{5}}\) - step5: Reduce fractions to a common denominator: \(\frac{26^{5}}{26^{5}}-\frac{6375600}{26^{5}}\) - step6: Transform the expression: \(\frac{26^{5}-6375600}{26^{5}}\) - step7: Reduce the fraction: \(\frac{742586-398475}{742586}\) - step8: Simplify: \(\frac{344111}{742586}\) ### Results of the Calculations #### Problem 15 The probability that at least two people choose the same letter when six people randomly select letters from the alphabet is approximately: \[ P(\text{at least one same}) \approx 0.4634 \quad \text{or} \quad \frac{344111}{742586} \] #### Problem 16 **a)** The probability of spelling "FOUR" in the correct order when tiles are replaced after each choice is: \[ P \approx 2.1883 \times 10^{-6} \quad \text{or} \quad \frac{1}{26^4} \] **b)** The probability of spelling "FOUR" in the correct order when tiles are not replaced after each choice is: \[ P \approx 2.7871 \times 10^{-6} \quad \text{or} \quad \frac{1}{358800} \] ### Summary of Answers 1. **Problem 14:** - a) \( P = \frac{1}{20} \) - b) (Interpretation needed for specific order) 2. **Problem 15:** - Probability that at least two choose the same letter: \( P \approx 0.4634 \) 3. **Problem 16:** - a) Probability with replacement: \( P \approx 2.1883 \times 10^{-6} \) - b) Probability without replacement: \( P \approx 2.7871 \times 10^{-6} \) If you have any further questions or need additional clarifications, feel free to ask!