Ayesha tosses a fair coin repeatedly until she gets a head. After she stops, Balu takes the coin and tosses it as many times as Ayesha had tossed it. What is the probability, computed up to three places after decimal, that Balu will also get exactly one head?

Let us analyze the problem step by step:

1. **Ayesha's Tosses**: 
   - Ayesha tosses a fair coin repeatedly until she gets a head. The probability of getting a head on any given toss is \( \frac{1}{2} \).
   - The number of tosses Ayesha makes follows a **geometric distribution** because she continues tossing until she gets her first head.
   - If Ayesha makes \( n \) tosses, then she must have gotten \( n - 1 \) tails followed by 1 head on the \( n \)-th toss. The probability that Ayesha makes exactly \( n \) tosses is:
     \[
     P(\text{Ayesha tosses } n \text{ times}) = \left(\frac{1}{2}\right)^{n}.
     \]

2. **Balu's Tosses**: 
   - Balu tosses the coin exactly \( n \) times (the same number of times Ayesha tossed).
   - The probability that Balu gets **exactly one head** in \( n \) tosses is the binomial probability of getting 1 head in \( n \) tosses. This is given by:
     \[
     P(\text{Balu gets exactly 1 head in } n \text{ tosses}) = \binom{n}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{n-1} = n \left(\frac{1}{2}\right)^n.
     \]

3. **Total Probability**: 
   - To find the overall probability that Balu also gets exactly one head, we sum over all possible values of \( n \) (i.e., the number of tosses Ayesha could have made). The total probability is:
     \[
     P(\text{Balu gets exactly 1 head}) = \sum_{n=1}^{\infty} P(\text{Ayesha tosses } n \text{ times}) \cdot P(\text{Balu gets exactly 1 head in } n \text{ tosses}).
     \]
   - Substituting the probabilities derived above, we get:
     \[
     P(\text{Balu gets exactly 1 head}) = \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n \cdot n \left(\frac{1}{2}\right)^n = \sum_{n=1}^{\infty} n \left(\frac{1}{2}\right)^{2n}.
     \]

4. **Simplifying the Series**:
   - We now need to evaluate the sum:
     \[
     S = \sum_{n=1}^{\infty} n \left(\frac{1}{4}\right)^n.
     \]
   - This is a standard power series, and the formula for such a series is given by:
     \[
     \sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2},
     \]
     where \( |x| < 1 \). In our case, \( x = \frac{1}{4} \), so we substitute this into the formula:
     \[
     S = \frac{\frac{1}{4}}{\left(1 - \frac{1}{4}\right)^2} = \frac{\frac{1}{4}}{\left(\frac{3}{4}\right)^2} = \frac{\frac{1}{4}}{\frac{9}{16}} = \frac{4}{9}.
     \]

Thus, the probability that Balu gets exactly one head is \( \frac{4}{9} \).

To express this as a decimal to three places:
\[
\frac{4}{9} \approx 0.444.
\]

Therefore, the probability that Balu will also get exactly one head is **0.444**.

Do you need more details or have any questions? Here are 5 related questions:

1. What is a geometric distribution and how is it used in this problem?
2. How is the binomial distribution used to model Balu’s tosses?
3. What is the general formula for summing a series involving powers of \( x \)?
4. How would the result change if the coin was biased instead of fair?
5. Can you derive the formula for the sum \( \sum_{n=1}^{\infty} n x^n \)?

**Tip**: In probability, geometric distributions are commonly used to model the number of trials until the first success in repeated independent Bernoulli trials (like coin tosses).

Discover the probability that Balu will get exactly one head after tossing a coin as many times as Ayesha, who tosses a fair coin until getting her first head. This problem involves geometric and binomial distributions, series summation, and advanced probability concepts. Ideal for students in Grades 11-12.

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