ISRO EC 2025 Q.21: solution and answer

Official answer: ASource: ISRO's final answer key of 15 May 2026

1. In a Poisson process of rate $\lambda$, the time $T$ between successive arrivals is exponentially distributed with density $f(t)=\lambda e^{-\lambda t}$, $t\ge 0$.

2. The mean arrival rate is $\lambda =10$ per hour.

3. Express the time in the same unit as the rate: $10\text{min}=\frac{1}{6}\text{h}$.

4. The survival function of the exponential distribution is $P(T>t)={\int }_t^{\infty }\lambda e^{-\lambda s}ds=e^{-\lambda t}$.

5. $$P\left(T>\frac{1}{6}\right)=\exp \left(-10\cdot \frac{1}{6}\right)=\exp \left(-\frac{10}{6}\right)\approx 0.189$$ This is option (a), $\exp (-10/6)\approx 0.188$ (the value $0.1889$ truncated to three places).

6. Option (b) has $\lambda t=1$, which would need a rate of 6 per hour. Options (c) and (d) have $\lambda t=5/6$, which corresponds to a rate of 5 per hour (or a mix-up of units); option (d) is simply $1$ minus option (c).