Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
To determine the number of times a fair coin must be thrown to achieve a probability of [tex]\(\frac{1}{64}\)[/tex] that every throw results in heads, follow these steps:
1. Understanding the Probability of a Coin Toss:
- A fair coin has two outcomes: heads or tails.
- The probability of getting heads on any single throw is [tex]\(\frac{1}{2}\)[/tex].
2. Setting Up the Problem:
- We are given that the probability of getting heads on every single throw, for a certain number of throws, is [tex]\(\frac{1}{64}\)[/tex].
3. Expressing the Probability Mathematically:
- Let [tex]\( n \)[/tex] be the number of times the coin is thrown.
- The probability of getting heads [tex]\( n \)[/tex] times in a row is [tex]\(\left(\frac{1}{2}\right)^n\)[/tex].
4. Equating the Given Probability:
- We know that:
[tex]\[ \left(\frac{1}{2}\right)^n = \frac{1}{64} \][/tex]
5. Solving for [tex]\( n \)[/tex]:
- Since [tex]\(\frac{1}{64}\)[/tex] can also be expressed as [tex]\(2^{-6}\)[/tex]:
[tex]\[ \left(\frac{1}{2}\right)^n = 2^{-6} \][/tex]
- Therefore, we equate the exponents of 2:
[tex]\[ \frac{1}{2}^n = 2^{-6} \Rightarrow n = 6 \][/tex]
6. Conclusion:
- So, the number of times the coin is thrown is [tex]\( n = 6 \)[/tex].
Therefore, a fair coin must be thrown 6 times to achieve a probability of [tex]\(\frac{1}{64}\)[/tex] that every throw results in heads.
1. Understanding the Probability of a Coin Toss:
- A fair coin has two outcomes: heads or tails.
- The probability of getting heads on any single throw is [tex]\(\frac{1}{2}\)[/tex].
2. Setting Up the Problem:
- We are given that the probability of getting heads on every single throw, for a certain number of throws, is [tex]\(\frac{1}{64}\)[/tex].
3. Expressing the Probability Mathematically:
- Let [tex]\( n \)[/tex] be the number of times the coin is thrown.
- The probability of getting heads [tex]\( n \)[/tex] times in a row is [tex]\(\left(\frac{1}{2}\right)^n\)[/tex].
4. Equating the Given Probability:
- We know that:
[tex]\[ \left(\frac{1}{2}\right)^n = \frac{1}{64} \][/tex]
5. Solving for [tex]\( n \)[/tex]:
- Since [tex]\(\frac{1}{64}\)[/tex] can also be expressed as [tex]\(2^{-6}\)[/tex]:
[tex]\[ \left(\frac{1}{2}\right)^n = 2^{-6} \][/tex]
- Therefore, we equate the exponents of 2:
[tex]\[ \frac{1}{2}^n = 2^{-6} \Rightarrow n = 6 \][/tex]
6. Conclusion:
- So, the number of times the coin is thrown is [tex]\( n = 6 \)[/tex].
Therefore, a fair coin must be thrown 6 times to achieve a probability of [tex]\(\frac{1}{64}\)[/tex] that every throw results in heads.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.