I need to find the probability that a normally-distributed random variable is greater than a uniformly-distributed random variable. That is:

\(X \sim \mathcal{N}(\mu, \sigma^2)\)

\(Y \sim \mathcal{U}(0, 1)\)

What I'm looking for is:

\(Pr(X \ge Y)\).

I got this thus far, but I'm not sure it's right, and it's not too useful in seeing how changing the mean and standard deviation affect the probability.

\(Pr(X \ge Y) = 1 - \int^1_0 \Phi(\frac{x-\mu}{\sigma}) dx\)

Is my solution on the right track? And is there an easier way to compute this?

\(X \sim \mathcal{N}(\mu, \sigma^2)\)

\(Y \sim \mathcal{U}(0, 1)\)

What I'm looking for is:

\(Pr(X \ge Y)\).

I got this thus far, but I'm not sure it's right, and it's not too useful in seeing how changing the mean and standard deviation affect the probability.

\(Pr(X \ge Y) = 1 - \int^1_0 \Phi(\frac{x-\mu}{\sigma}) dx\)

Is my solution on the right track? And is there an easier way to compute this?

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