Answer :
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.
