GATE EE 2023 Set 1 — Question 45

Let $p$ be the probability of getting a 'head' (H) in a single trial of the biased coin. The number of trials required for the first occurrence of a head follows a Geometric distribution.The expected number of trials, $E[X]$, for the first success in a series of Bernoulli trials is given by:$$E[X] = \frac{1}{p}$$Given that the expected number of trials for the first occurrence of a head is 4, we have:$$E[X] = 4 = \frac{1}{p}$$Solving for $p$, we get the probability of getting a head:$$p = \frac{1}{4} = 0.25$$The probability of not getting a head (i.e., getting a 'tail', T) is $q = 1 - p$:$$q = 1 - 0.25 = 0.75$$The question asks for the probability of the first occurrence of a head in the second trial. This means the first trial must be a tail (T) and the second trial must be a head (H). The sequence of outcomes is TH.The probability of this event is given by:$$P(\text{first head on 2nd trial}) = P(T) \times P(H) = q \times p$$Substituting the values of $p$ and $q$:$$P(\text{first head on 2nd trial}) = 0.75 \times 0.25 = 0.1875$$Alternatively, using the probability mass function of the Geometric distribution, $P(X=k) = (1-p)^{k-1}p$. For $k=2$:$$P(X=2) = (1 - 0.25)^{2-1} \times 0.25 = (0.75)^1 \times 0.25 = 0.1875$$The question asks to round the answer to 3 decimal places.$$0.1875 \approx 0.188$$Thus, the probability is 0.188.