Probability \(p\): At Least One Tail

The problem involves calculating probabilities related to tossing an unbiased coin \(n\) times. We are given definitions for two probabilities:

• \(p\): Represents the probability of obtaining at least one tail.
• \(q\): Represents the probability of obtaining at least two tails.

To find \(p\), we consider the complementary event: getting no tails at all, which means getting all heads. Since the coin is unbiased, the probability of getting heads on a single toss is \(\frac{1}{2}\). For \(n\) tosses, the probability of getting all heads is \((\frac{1}{2})^n\).

Therefore, the probability of getting at least one tail, \(p\), is:

\(p = 1 - P(\text{all heads}) = 1 - (\frac{1}{2})^n\)

Probability \(q\): At Least Two Tails

To find \(q\), the probability of getting at least two tails, we consider the complementary event: getting zero tails or exactly one tail.

The probability of getting zero tails (\(P(T_0)\)) is the same as getting all heads, which is \((\frac{1}{2})^n\).

The probability of getting exactly one tail (\(P(T_1)\)) can be calculated using the binomial probability formula. There are \(\binom{n}{1}\) ways to choose the position for the single tail, and the probability for each specific sequence (like T H H...) is \((\frac{1}{2})^n\).

\(P(T_1) = \binom{n}{1} (\frac{1}{2})^n = n (\frac{1}{2})^n\)

So, the probability \(q\) is:

\(q = 1 - [P(\text{zero tails}) + P(\text{one tail})]\) \(q = 1 - [(\frac{1}{2})^n + n (\frac{1}{2})^n]\)

Factoring out \((\frac{1}{2})^n\):

\(q = 1 - (\frac{1}{2})^n (1+n)\)

Condition Analysis: \(p-q = \frac{5}{32}\)

We are given the equation \(p - q = \frac{5}{32}\). Substitute the derived expressions for \(p\) and \(q\):

\([1 - (\frac{1}{2})^n] - [1 - (\frac{1}{2})^n (1+n)] = \frac{5}{32}\)

Now, simplify the equation by removing the brackets and combining terms:

\(1 - (\frac{1}{2})^n - 1 + (\frac{1}{2})^n (1+n) = \frac{5}{32}\)

The '1's cancel out:

\((\frac{1}{2})^n (1+n) - (\frac{1}{2})^n = \frac{5}{32}\)

Factor out the common term \((\frac{1}{2})^n\):

\((\frac{1}{2})^n [(1+n) - 1] = \frac{5}{32}\)

Simplify the expression inside the brackets:

\((\frac{1}{2})^n [n] = \frac{5}{32}\)

This can be written as:

\(\frac{n}{2^n} = \frac{5}{32}\)

To find the value of \(n\), we can test small positive integer values since \(n\) represents the number of tosses:

• If \(n=1\), \(\frac{1}{2^1} = \frac{1}{2} \neq \frac{5}{32}\)
• If \(n=2\), \(\frac{2}{2^2} = \frac{2}{4} = \frac{1}{2} \neq \frac{5}{32}\)
• If \(n=3\), \(\frac{3}{2^3} = \frac{3}{8} \neq \frac{5}{32}\)
• If \(n=4\), \(\frac{4}{2^4} = \frac{4}{16} = \frac{1}{4} \neq \frac{5}{32}\)
• If \(n=5\), \(\frac{5}{2^5} = \frac{5}{32}\). This matches the equation.

So, the number of coin tosses is \(n=5\).

Value Calculation: \(p+q\)

With \(n=5\), we can now calculate the specific values for \(p\) and \(q\):

\(p = 1 - (\frac{1}{2})^5 = 1 - \frac{1}{32} = \frac{31}{32}\)

\(q = 1 - (\frac{1}{2})^5 (1+5) = 1 - (\frac{1}{32})(6) = 1 - \frac{6}{32} = \frac{26}{32} = \frac{13}{16}\)

The question asks for the value of \(p + q\).

\(p + q = \frac{31}{32} + \frac{13}{16}\)

To add these fractions, we need a common denominator, which is 32. Convert \(\frac{13}{16}\) to an equivalent fraction with a denominator of 32:

\(\frac{13}{16} = \frac{13 \times 2}{16 \times 2} = \frac{26}{32}\)

Now, add the fractions:

\(p + q = \frac{31}{32} + \frac{26}{32} = \frac{31 + 26}{32}\)

\(p + q = \frac{57}{32}\)

Therefore, the value of \(p + q\) is \(\frac{57}{32}\).