Question

The expected number of trials for first occurrence of a "head" in a biased coin is known to be 4. The probability of first occurrence of a "head" in the second trial is .................... (Round off to 3 decimal places).

Hint:

In geometric distribution, use \(P(N=k) = (1-p)^{k-1} p\). The expected value gives you \(p\) directly as \(1/E[N]\).

Answer 1

Step 1: Expected value condition.
For a geometric distribution, the expected number of trials until the first success is: \[ E[N] = \frac{1}{p} \] where \(p\) is the probability of head. Given \(E[N] = 4\): \[ \frac{1}{p} = 4 \Rightarrow p = 0.25 \]

Step 2: Probability of first head on 2nd trial.
For geometric distribution: \[ P(\text{first head on 2nd trial}) = (1-p)^{2-1} \cdot p \] \[ = (1 - 0.25)(0.25) = (0.75)(0.25) = 0.1875 \]

Final Answer:
\[ \boxed{0.188} \]