What is the probability that Alex applies for his first sick leave on the fifth day? Alex will work on a consulting project for SALT Solutions for $5$ days.

During these $5$ days, the probability that Alex applies for sick leave on a particular day is the same, which is less than $0.5$.

The probability that Alex applies for his first sick leave on the second day is $0.21$.

The event that Alex applies for sick leave on a particular day is independent of the event that Alex applies for sick leave on other days.

What is the probability that Alex applies for his first sick leave on the fifth day?



What in the world is this question even saying? What does it mean that the probability that Alex applies for sick leave on a particular day is the same, which is less than $.5$? Is this saying that the probability for applying for each sick day is $.21$ independent of every other day since he does that on the second day? If so, is this just a geometric random variable with $p = .21$ so the answer is $$.21(1-.21)^4?$$
You are correct in saying that it is a geometric random variable. The first part where it says that the probability that he applies for sick leave on any of his days is the same simply satisfies the requirement for geometric distributions that the probability of the outcome occurring on each trial is the same (you can ignore the part that says that it is less than 0.5 since the question later gives you the exact probability of success which is 0.21). Since the probability of him applying for his first sick leave on any day is the same it means that the probability of him applying for sick leave on day "n" is the same as him applying for sick leave on day 2 which equals 0.21. It all really boils down to this being a tricky or poorly written question.
Day 2 means that the person didn't get sick on day one but got sick on day = .21 = p(1-p) => Quadratic Equation solve for p. p=.3 q=.7 
That is why it states that p is less than .5

Solution 

(.3).70^4 = .07203