Answer: 0.25
Step-by-step explanation:
Given : The future lifetime of Gary is uniformly distributed with interval [0 years , 60 years].
Then the probability density function for Gary's future lifetime will be:-
[tex]f(x)=\dfrac{1}{60-0}=\dfrac{1}{60}[/tex]
The future lifetime of Eric is uniformly distributed with interval [0 years , 40 years].
Then the probability density function for Erin's future lifetime will be:-
[tex]f(x)=\dfrac{1}{40-0}=\dfrac{1}{40}[/tex]
Now, the joint density function for Gary and Eric's future lifetime :-
[tex]f(x,y)=f(x)f(y)=\dfrac{1}{40\times60}=\dfrac{1}{2400}[/tex] [∵Their future lifetimes are independent. ]
Now, the probability that Gary dies first is given by :-
[tex]\int^{60}_{0}\int^{40}_{x}f(x,y)\ dy\ dx\\\\=\int^{60}_{0}\int^{40}_{x}\dfrac{1}{2400}\ dy\ dx\\\\=\int^{60}_{0}\dfrac{40-x}{2400}\ dx\\\\=\dfrac{1}{2400}[40x-\dfrac{x^2}{2}]^{60}_{0}\\\\=\dfrac{1}{2400}(2400-\dfrac{3600}{2})=0.25[/tex]
Hence, the probability that Gary dies first =0.25
