Probability that x is greater than some value and y is less than some value where values come from normal distribution? I was attempting the following self-study question. After repeated tries, my answers were at best 0.0944, however, the model answer seemed to be otherwise at 0.1889.
Will appreciate any pointers as I simply can't figure out even after repeated attempts.
Question

According to the 2010 census in a city, university-educated employee earned a mean annual income of USD 61,000 & standard deviation of USD4,000. If the incomes are normally distributed and two university-educated employees are randomly selected, what is the probability that one of the them earns more than USD66,000 per year and the other earns less than USD66,000 per year?

My Attempt

 A: It is a TRUE/FALSE fact to earn more than USD66,000 per year and you correctly calculated that the probability for a university-educated employee to be one of those people (probability of TRUE) is 0.1056. Now, among the two person you select you should have one TRUE statement. To formalize that situation let X define the number of TRUE statement among the n = 2 repetitions of an experiment with probability 0.1056. You should calculate P(X = 1). The rest should be clear.
Update: your answer would be correct if the position of TRUE statement was fixed in the first or the second selected employee. Since no such restriction applies there are more than one combinations and that is the reason that binomial distribution should be applied.
A: You should apply the "Law of total probability" to get the correct answer (0.1889).
Your problem is considering two random employees whose salaries (X and Y) are normally distributed with mean 61000 and st. dev. 4000.
The probability you have calculated is: $$P(Y > 66000 \cap X < 66000)$$ You have to sum it to $P(X > 66000 \cap Y < 66000)$ to get: $$P[(X > 66000 \cap Y < 66000) \cup (Y > 66000 \cap X < 66000)]$$ that is the correct answer of your problem (0.1889).
