Question

You are tossing a fair coin repeatedly until a ‘Head’ appears. The expected number of tosses required for a ‘Head’ to appear is ...........

Hint:

For a geometric distribution with success probability \( p \), the expected number of trials to get the first success is \( E(X) = \frac{1}{p} \).

Answer 1

Let \( X \) be the random variable denoting the number of tosses required to get the first ‘Head’. We are asked to find the expected value \( E(X) \) of \( X \).
Step 1: Since the coin is fair, the probability of getting a ‘Head’ on each toss is \( P(\text{Head}) = \frac{1}{2} \). The random variable \( X \) follows a geometric distribution with parameter \( p = \frac{1}{2} \).
Step 2: The expected value for a geometric distribution is given by: \[ E(X) = \frac{1}{p} \] Step 3: Substituting \( p = \frac{1}{2} \): \[ E(X) = \frac{1}{\frac{1}{2}} = 2 \] Thus, the expected number of tosses required to get a ‘Head’ is 2. Final Answer: \[ \boxed{2} \]