An unbiased coin is tossed $n$ times. The probability of getting at least one tail is $p$ and the probability of at least two tails is $q$ and $p-q = \frac{5}{32}$.
Understanding Coin Toss Probabilities
The problem involves calculating the value of '\(n\)', the number of times an unbiased coin is tossed. We are given information about the probabilities of two events: getting at least one tail and getting at least two tails.
- An unbiased coin means the probability of getting a head (H) is equal to the probability of getting a tail (T), i.e., \(P(H) = P(T) = \frac{1}{2}\).
- When the coin is tossed \(n\) times, the total number of possible outcomes is \(2^n\).
Calculating Probability 'p'
Let '\(p\)' be the probability of getting at least one tail.
The complementary event to "at least one tail" is "getting no tails", which means getting all heads.
The probability of getting all heads in \(n\) tosses is \(P(\text{all H}) = \left(\frac{1}{2}\right)^n\).
Therefore, the probability of getting at least one tail is:
\(p = P(\text{at least one T}) = 1 - P(\text{no T}) = 1 - P(\text{all H}) = 1 - \left(\frac{1}{2}\right)^n\)
Calculating Probability 'q'
Let '\(q\)' be the probability of getting at least two tails.
The complementary event to "at least two tails" is "getting zero tails (all heads)" or "getting exactly one tail".
- Probability of getting zero tails (all heads) is \(P(\text{0 T}) = \left(\frac{1}{2}\right)^n\).
- Probability of getting exactly one tail: This can occur in \(n\) different ways (e.g., T H H..., H T H..., H H T...). Each specific sequence has a probability of \(\left(\frac{1}{2}\right)^n\). So, \(P(\text{exactly 1 T}) = n \times \left(\frac{1}{2}\right)^n\).
The probability of getting at least two tails is:
\(q = P(\text{at least 2 T}) = 1 - [P(\text{0 T}) + P(\text{exactly 1 T})]\)
\(q = 1 - \left[\left(\frac{1}{2}\right)^n + n \times \left(\frac{1}{2}\right)^n\right]\)
\(q = 1 - \left(\frac{1}{2}\right)^n \times (1 + n)\)
Using the Given Equation to Find 'n'
We are given the equation: \(p - q = \frac{5}{32}\).
Substitute the expressions for \(p\) and \(q\):
\(\left[1 - \left(\frac{1}{2}\right)^n\right] - \left[1 - (1+n) \times \left(\frac{1}{2}\right)^n\right] = \frac{5}{32}\)
Simplify the equation:
\(1 - \left(\frac{1}{2}\right)^n - 1 + (1+n) \times \left(\frac{1}{2}\right)^n = \frac{5}{32}\)
Combine terms involving \(\left(\frac{1}{2}\right)^n\):
\((1+n) \times \left(\frac{1}{2}\right)^n - \left(\frac{1}{2}\right)^n = \frac{5}{32}\)
Factor out \(\left(\frac{1}{2}\right)^n\):
\(\left(\frac{1}{2}\right)^n \times [(1+n) - 1] = \frac{5}{32}\)
\(\left(\frac{1}{2}\right)^n \times n = \frac{5}{32}\)
This simplifies to:
\(\frac{n}{2^n} = \frac{5}{32}\)
Solving for 'n' by Testing Values
We need to find the integer value of '\(n\)' that satisfies the equation \(\frac{n}{2^n} = \frac{5}{32}\). We can test the given options:
- For \(n=4\): \(\frac{4}{2^4} = \frac{4}{16} = \frac{1}{4}\). This is not equal to \(\frac{5}{32}\).
- For \(n=5\): \(\frac{5}{2^5} = \frac{5}{32}\). This matches the required value.
- For \(n=6\): \(\frac{6}{2^6} = \frac{6}{64} = \frac{3}{32}\). This is not equal to \(\frac{5}{32}\).
- For \(n=7\): \(\frac{7}{2^7} = \frac{7}{128}\). This is not equal to \(\frac{5}{32}\).
The only value of \(n\) from the options that satisfies the equation is \(n=5\).
Conclusion
The value of \(n\) for the given coin toss probabilities is 5.
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