On the one hand, it would seem that since the probability is the same for rain on each day, the chance he will return on day one two or three is 1/3 or .3333. On the other hand, since the probability of rain is given as 20%, it makes sense to say that by day three, the probability that he will head home each day is the same as the probability it will rain on that day. Where it gets complicated is when you start compounding probabilities that it will or won't rain on each day. For him to be headed home on day three, then it must not have rained on days one and two. The probability he heads home on day three is therefore the probability that it did NOT rain on day one, times the probability it did NOT rain on day two times the probability it DID RAIN on DAY three, or (.8 *.8*.2) or 12.8%. That's my guess and I'm sticking to it, so to all the mathematicians out there, please chime in! Is this right?

In probability, the day he returns can be thought of as a geometric random variable.

Monday is the third day.
The only way he will return on the third day is: no rain on the 1st day, no rain on the 2nd day, and rain on the 3rd day.

Since the probability that each day will rain can be assumed to be independent, then

P(return on 3rd day) = P(no rain on 1st) * P(no rain on 2nd) * P(rain on 3rd)
= 0.8 * 0.8 *0.2 = 0.128

I would agree with bryantmichaels.

this answer is right