Question: The first prize winner in a lottery has to guess 5 numbers in correct order from a list of 40 numbers.
The 2nd prize winner in a lottery has to guess 4/5 number in correct order rom a list of 40 numbers

What is the probability of winning the first OR second prize?

I know the odds of first prize winner are 1/40P5 and the second 1/40P4 but to get the odds of winning either what do I have to do with both odds? (multiply, add, subtract etc.) Help would be greatly appreciated =)

Answer: 
Let's see: $1/{^{40}\mathrm P\_5}$ is the probability of obtaining a specific tuple of five numbers, when selecting in order 5 from 40 numbers. Okay.

Similarly: $1/{^{40}\mathrm P\_4}$ is the probability of obtaining a specific tuple of four numbers, when selecting in order 5 from 40 numbers.

This would be okay if the second prize is awarded to people who have a match for the first four numbers they list.   *However* we would likely want to avoid giving second prize to anyone who also wins first prize.   So we need to exclude the probability of qualifying for second prize but winning first.

But wait...

Indeed, $1/{^{40}\mathrm P\_4}$ is the probability that the first four numbers you list match the first four balls as they drop in order, and *maybe* the fifth on the list also matches the fifth ball dropped.

That is: It is the answer you seek.   The probability of winning either first or second prize is $1/{^{40}\mathrm P\_4}$
Taking that those who are eligible for a first prize are excluded from a second prize,  

and that guessing, e.g. $\boxed{92345}\;$ or $\;\boxed{12347}$ against the correct $\boxed{12345}$ will win you a second prize,

P(win first prize) $= \dfrac{1}{^{40}P\_5} = \dfrac{1}{40}\dfrac1{39}\dfrac1{38}\dfrac1{37}\dfrac1{36}\;\;$ [ Put in this form to compare with second prize ]

The only ways to get just the second prize is to get either the first number XOR the last number not in the drawn $5$, with $4$ consecutive ones correct.

P(win second prize) $= \dfrac{35}{40}\dfrac1{39}\dfrac1{38}\dfrac1{37}\dfrac1{36} + \dfrac{1}{40}\dfrac1{39}\dfrac1{38}\dfrac1{37}\dfrac{35}{36}$

Since the cases are disjoint, add the two probabilities for the answer.