A program consists of two modules executed sequentially. Let f_1(t) and f_2(t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by A. f_1(t) + f_2(t) B. integral_{0}^{t} f_1(x) * f_2(x) dx C. integral_{0}^{t} f_1(x) * f_2(t - x) dx D. max{ f_1(t), f_2(t) }

GATE 2003 · Engineering Mathematics · Probability Density Function · medium

Answer: C. integral_{0}^{t} f_1(x) * f_2(t - x) dx

1. Set up the sum of two independent RVs: Let T_1 take value x with density f_1(x). Then T_2 = t - x with density f_2(t-x). Multiply and integrate over all valid x in [0, t].
2. Eliminate wrong options: Option A (sum) applies to mutually exclusive events, not sequential times. Option B integrates f_1(x)*f_2(x) — same argument for both, not correct. Option D (max) is for the maximum of two RVs, not their sum. Only option C has the convolution form f_1(x)*f_2(t-x).