3 A coin is loaded so that the probability of a head occurring on a single toss is In six tosses of the coin, what is the 4 probability of getting all heads or all tails? The probability of all heads or all tails is (Round to three decimal places as needed.) …
3 A coin is loaded so that the probability of a head occurring on a single toss is In six tosses of the coin, what is the 4 probability of getting all heads or all tails? The probability of all heads or all tails is (Round to three decimal places as needed.) …
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![**Problem Statement:**
A coin is loaded so that the probability of a head occurring on a single toss is \( \frac{3}{4} \). In six tosses of the coin, what is the probability of getting all heads or all tails?
**Solution Space:**
The probability of all heads or all tails is [Blank].
(Round to three decimal places as needed.)
---
**Explanation:**
To solve this problem, we need to calculate two separate probabilities and then add them together:
1. Probability of getting all heads in six tosses.
2. Probability of getting all tails in six tosses.
### Calculations:
- **Probability of all heads:**
The probability of getting one head is \( \frac{3}{4} \). Therefore, the probability of getting all heads in six tosses is:
\[
\left( \frac{3}{4} \right)^6
\]
- **Probability of all tails:**
The probability of getting one tail is \( 1 - \frac{3}{4} = \frac{1}{4} \). Therefore, the probability of getting all tails in six tosses is:
\[
\left( \frac{1}{4} \right)^6
\]
### Total Probability:
- Add both probabilities together to get the total probability of all heads or all tails:
\[
\left( \frac{3}{4} \right)^6 + \left( \frac{1}{4} \right)^6
\]
Finally, calculate the expression and round the result to three decimal places as required.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f970362-e46d-414b-99ae-39ca4f64716b%2F1ff8fa50-dcf6-434b-9cd3-a60d1ee288ec%2F3qd08rs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A coin is loaded so that the probability of a head occurring on a single toss is \( \frac{3}{4} \). In six tosses of the coin, what is the probability of getting all heads or all tails?
**Solution Space:**
The probability of all heads or all tails is [Blank].
(Round to three decimal places as needed.)
---
**Explanation:**
To solve this problem, we need to calculate two separate probabilities and then add them together:
1. Probability of getting all heads in six tosses.
2. Probability of getting all tails in six tosses.
### Calculations:
- **Probability of all heads:**
The probability of getting one head is \( \frac{3}{4} \). Therefore, the probability of getting all heads in six tosses is:
\[
\left( \frac{3}{4} \right)^6
\]
- **Probability of all tails:**
The probability of getting one tail is \( 1 - \frac{3}{4} = \frac{1}{4} \). Therefore, the probability of getting all tails in six tosses is:
\[
\left( \frac{1}{4} \right)^6
\]
### Total Probability:
- Add both probabilities together to get the total probability of all heads or all tails:
\[
\left( \frac{3}{4} \right)^6 + \left( \frac{1}{4} \right)^6
\]
Finally, calculate the expression and round the result to three decimal places as required.
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