The rote way to compute $P[Y>X]$ is by double integral

$$\int_0^\infty f_X(x) dx \int_x^\infty f_Y(y) dy $$

Where the inner integral may be recognized as the survival function of $Y$, an exponential with parameter $\lambda=1$, at $x$, equal to $e^{-x}$. Then the remaining integral

$$\int_0^\infty e^{-x} f_X(x) dx $$

may be recognized as the moment generating function of $X$ evaluated at $-1$. The MGF of a $\rm{Gamma}$ is $(1-\theta t)^{-k}$, which for $\theta = 3, k=3, t=-1$ is

$$(1+3)^{-3} = 0.015625$$

The question was for $P[X>Y] = 1-P[Y>X]$, so we want

$$1-(1+3)^{-3} = 1-0.015625 = 0.984375$$

which agrees with soakley's answer.

The rote way to compute $P[Y>X]$ is by double integral

$$\int_0^\infty f_X(x) dx \int_x^\infty f_Y(y) dy $$

Where the inner integral may be recognized as the survival function of $Y$, an exponential with parameter $\lambda=1$, at $x$, equal to $e^{-x}$. Then the remaining integral

$$\int_0^\infty e^{-x} f_X(x) dx $$

may be recognized as the moment generating function of $X$ evaluated at $-1$. The MGF of a $\rm{Gamma}$ is $(1-\theta t)^{-k}$, which for $\theta = 3, k=3, t=-1$ is

$$(1+3)^{-3} = 0.015625$$

The question was for $P[X>Y] = 1-P[Y>X]$, so we want

$$1-(1+3)^{-3} = 1-0.015625 = 0.984375$$

which agrees with soakley's answer.

The rote way to compute $P[Y>X]$ is by double integral

$$\int_0^\infty f_X(x) dx \int_x^\infty f_Y(y) dy $$

Where the inner integral, the survival function of $Y$, an exponential with parameter $\lambda=1$, is $e^{-x}$. We recognize that

$$\int_0^\infty e^{-x} f_X(x) dx $$

is the moment generating function of $Y$ evaluated at $-1$. The MGF of a $\rm{Gamma}$ is $(1-\theta t)^{-k}$, which for $\theta = 3, k=3, t=-1$ is

$$(1+3)^{-3} = 0.015625$$

The question was for $P[X>Y] = 1-P[Y>X]$, so we want

$$1-(1+3)^{-3} = 1-0.015625 = 0.984375$$

which agrees with soakley's answer.

The rote way to compute $P[Y>X]$ is by double integral

$$\int_0^\infty f_X(x) dx \int_x^\infty f_Y(y) dy $$

Where the inner integral may be recognized as the survival function of $Y$, an exponential with parameter $\lambda=1$, at $x$, equal to $e^{-x}$. Then the remaining integral

$$\int_0^\infty e^{-x} f_X(x) dx $$

may be recognized as the moment generating function of $X$ evaluated at $-1$. The MGF of a $\rm{Gamma}$ is $(1-\theta t)^{-k}$, which for $\theta = 3, k=3, t=-1$ is

$$(1+3)^{-3} = 0.015625$$

The question was for $P[X>Y] = 1-P[Y>X]$, so we want

$$1-(1+3)^{-3} = 1-0.015625 = 0.984375$$

which agrees with soakley's answer.