Forgetful parents

We need to consider the possibilities that all 5 get the correct colour, 3 get the correct colour, 2 get it and only 1 get his/her favourite colour. Note that the probability that exactly 1 get a wrong colour (i.e. 4 get their favourite colour) is zero as there must be a mix up between at least 2 people in order for there to be an error.

Case 1: All 5 get the correct colour.

There is only 1 possibility.

Case 2: Exactly 3 people get the correct colour.

We only need to consider how many possibilities are there for any 2 to get the wrong colour. Obviously, $5C2 = 10$ is the answer as there is only 1 way to mix up the colours between 2 people aso we only need to consider which 2 have their colours mixed up.

Case 3: Exactly 2 people get the correct colour.

In considering how many ways are there for 3 people to get the wrong colour, what we do is to first consider a group of 3 people. Here, the number of possibilities is n(total) - n(all correct) - n(2 get wrong colour) = $3! - 1 -3C2 = 6-1 -3=2$ where n(event) is defined as the number of ways for the event to occur. Remember that exactly 1 person getting a wrong colour is absurd. Now expanding this context to a group of 5, we have $2 \times 5C3 =2 \times 10 = 20$ ways over here as we also need to consider which 3 of the 5 get a wrong colour.

Case 4: Exactly 1 gets the correct colour.

Like what we did in Case 3, consider how many ways are there for all 4 people in a group to get the wrong colour.We have n(total) - n(all correct) - n(2 get the wrong colour) - n(3 get the wrong colour) = $4! - 1 - 4C2 - 2 \times 4C3 = 24 - 1- 6 - 2 \times 4 = 9$ . Expanding this to the context of 5 people, we have $9 \times 5C4 = 9 \times 5 = 45$ . As you can see here, in calculating n(3 get the wrong colour), we need to expand the context from 3 to 4 people, hence $2 \times 4C3$ as we need to choose 3 out of 4.

Finally, we do n(all wrong) = n(total) - n(all correct) - n(2 wrong) - n(3 wrong) - n(4 wrong) = $5! - 1 - 10 -20-45 = 120-76 = 44$ .

We have $\frac{44}{120} = \frac{11}{30}$ so $a+b = 11+30 = \boxed{41}$