Let's say A is a normal distribution A = N(mean = 0, SD = 1). Let's say B = A + N(0, 0.5), such that A and B are dependent. What is P(A>1, B>1) (the probability of seeing a point whose A and B values are both >1)?

Apologies if my notation isn't quite right, my pure math background isn't that strong. I'm currently trying to understand some concepts about dependence and joint probability.

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    $\begingroup$ Let me get this straight: are you looking for finding the joint distribution from marginals? $\endgroup$ Dec 14, 2022 at 10:50
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    $\begingroup$ @User1865345, I would understand the question to mean that $B = A+C$, where $C \sim N(0, 0.5)$ is independent of $A$. Then it would be a well-posed question. (Assuming, of course, that that's what is meant.) $\endgroup$ Dec 14, 2022 at 14:18
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    $\begingroup$ I see @JukkaKohonen. It's just that the language opted by OP seems to be ambiguous. $\endgroup$ Dec 14, 2022 at 14:22
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    $\begingroup$ I presume you intend that the variable you add to $A$ to get $B$ is intended to be independent of $A$. You should be explicit. If that is the case, you would normally say something like "Let $A$ and $D$ be independent random variables where $A$ has a standard normal distribution and $D$ has mean $0$ and standard deviation $0.5$. Let $B=A+D$. What is $P(A>1, B>1)$? ... If that is the case I believe this may be a duplicate. $\endgroup$
    – Glen_b
    Dec 14, 2022 at 17:17
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    $\begingroup$ stats.stackexchange.com/questions/583006 contains a general answer. It applies because $\Pr(A\gt 1,B\gt 1)=\Pr(-A\lt -1,-B\lt -1)=\Pr(-A\le 1,-B\le 1)$ is, by definition, the value of the CDF of $(-A,-B)$ at the point $(-1,-1)$ and $(-A,-B)$ has a bivariate Normal distribution provided the "N(0,0.5)" represents some variable $D$ where $(A,D)$ has a bivariate Normal distribution (which is automatically the case when, as @Glen_b points out, $D$ is independent of $A$). It's also a special case of stats.stackexchange.com/questions/213456 where $a=-\infty.$ $\endgroup$
    – whuber
    Dec 14, 2022 at 19:14