Answer:
[tex]\text{The probability that Gary dies first = }\frac{1}{3}[/tex]
Step-by-step explanation:
To find the required probability,
We can split the 40 years of Eric into very small pieces of time dt. Now, If Eric dies at time t and Gary dies before Eric, the probability is :
[tex]\frac{t}{60}\times \frac{dt}{40}=\frac{t\cdot dt}{2400}[/tex]
Hence the probability that Gary dies first is given by :
[tex]=\int_{0}^{40}\frac{t}{2400}dt \\\\= \frac{1}{3}[/tex]
