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QUESTION 17 Let's suppose someone flips a fair coin three times. 1. How many outcomes are possible? 2. What is the probability of getting exactly one tail? Give as a fraction 3. What is the probability of getting at most two tails? Give asa fraction

QUESTION 17 Let's suppose someone flips a fair coin three times. 1. How many outcomes are possible? 2. What is the probability of getting exactly one tail? Give as a fraction 3. What is the probability of getting at most two tails? Give asa fraction

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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### Probability Questions for Educational Practice #### Coin Flipping Probability **Question 17:** Let's suppose someone flips a fair coin three times. 1. **How many outcomes are possible?** - There are 2 outcomes for each coin flip (Heads or Tails). Therefore, flipping a coin three times results in \(2^3 = 8\) possible outcomes. 2. **What is the probability of getting exactly one tail? Give as a fraction.** - The possible outcomes with exactly one tail are: HTT, THT, TTH. There are 3 such outcomes. Since there are 8 possible outcomes in total, the probability is \(\frac{3}{8}\). 3. **What is the probability of getting at most two tails? Give as a fraction.** - Getting at most two tails means getting 0, 1, or 2 tails. The possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH. There are 7 favorable outcomes. The probability is \(\frac{7}{8}\). #### Dice Rolling Probability **Question 18:** Consider the following experiment: roll a pair of fair dice and observe the sample space. Answer the following questions based on the above sample space. 1. **What is the probability of obtaining a total score (sum of the faces of both dice) of 11?** - The possible outcomes with a total score of 11 are (5,6) and (6,5), which are 2 favorable outcomes. Since there are 36 possible outcomes when rolling a pair of dice (6 sides each, so \(6 \times 6 = 36\)), the probability is \(\frac{2}{36} = \frac{1}{18}\). 2. **What is the probability of obtaining a total score of 9?** - The possible outcomes with a total score of 9 are (3,6), (4,5), (5,4), and (6,3), which are 4 favorable outcomes. Therefore, the probability is \(\frac{4}{36} = \frac{1}{9}\). **Instructions:** Click "Save and Submit" to save and submit your answers. Click "Save All Answers" to save all answers. ### Weather Information: - 74°F
Transcribed Image Text:### Probability Questions for Educational Practice #### Coin Flipping Probability **Question 17:** Let's suppose someone flips a fair coin three times. 1. **How many outcomes are possible?** - There are 2 outcomes for each coin flip (Heads or Tails). Therefore, flipping a coin three times results in \(2^3 = 8\) possible outcomes. 2. **What is the probability of getting exactly one tail? Give as a fraction.** - The possible outcomes with exactly one tail are: HTT, THT, TTH. There are 3 such outcomes. Since there are 8 possible outcomes in total, the probability is \(\frac{3}{8}\). 3. **What is the probability of getting at most two tails? Give as a fraction.** - Getting at most two tails means getting 0, 1, or 2 tails. The possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH. There are 7 favorable outcomes. The probability is \(\frac{7}{8}\). #### Dice Rolling Probability **Question 18:** Consider the following experiment: roll a pair of fair dice and observe the sample space. Answer the following questions based on the above sample space. 1. **What is the probability of obtaining a total score (sum of the faces of both dice) of 11?** - The possible outcomes with a total score of 11 are (5,6) and (6,5), which are 2 favorable outcomes. Since there are 36 possible outcomes when rolling a pair of dice (6 sides each, so \(6 \times 6 = 36\)), the probability is \(\frac{2}{36} = \frac{1}{18}\). 2. **What is the probability of obtaining a total score of 9?** - The possible outcomes with a total score of 9 are (3,6), (4,5), (5,4), and (6,3), which are 4 favorable outcomes. Therefore, the probability is \(\frac{4}{36} = \frac{1}{9}\). **Instructions:** Click "Save and Submit" to save and submit your answers. Click "Save All Answers" to save all answers. ### Weather Information: - 74°F
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