# 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

• Think about binomial distribution and notice that your answer is exactly the half of the correct answer. Update with another hint: C(2, 1) = 2!/1!1! = 2. Commented May 4, 2014 at 7:39
• @Epaminondas thanks for the hint. I see that with 2C1 = 2, if we multiply 0.0944 by 2, we will get the answer 0.1889. However, what I am unclear still is how does binomial come into play in this question where we are both looking for failure and success at the same time? Commented May 4, 2014 at 10:36

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).