GATE 2023 EE Q.45: solution and answer

Official answer: 0.187 to 0.188Source: Official answer key, GATE 2023 EE (IIT Kanpur)

1. Let $p$ be the probability of a head on a single toss and $q=1-p$ the probability of a tail. The tosses are independent, so the number of trials $X$ up to and including the first head has a geometric distribution: $$P(X=k)=q^{k-1}p,k=1,2,3,\dots$$

2. Mean of this distribution: $$E[X]=\sum _{k=1}^{\infty }k{q}^{k-1}p=\frac{p}{(1-q{)}^2}=\frac{1}{p}$$

3. We are told $E[X]=4$, so $\frac{1}{p}=4\Rightarrow p=\frac{1}{4}=0.25$, and the probability of a tail is $q=1-p=0.75$.

4. For the first head to come on the second trial, the first toss must be a tail and the second a head: $$P(X=2)=qp=(0.75)(0.25)=\frac{3}{16}=0.1875$$

5. Rounded to 3 decimal places: $P(X=2)\approx 0.188$. (The exact value $0.1875$ sits right on the rounding boundary, which is why the official key accepts anything from $0.187$ to $0.188$.)