What is the probability of getting at least one head if you flip a coin 10 times

some simple equations and trees to help explain this would be great, thank you

A common trick is to invert so that you calculate something easier, then subtract from 1.

Here, P(10 tails)=(1/2)^10 = 1/1024

So, P(at least one head)=1 - 1/1024 = 1023/1024

Whenever you are asked to calculate probabilities like "at least one" then you should see that as a big hint to calculate "none", then invert it. If the probability of none is x, then the probability of at least one is 1-x.

You could go the hard way and add up the probability of exactly one, then exactly two, three etc. and use diagrams and trees of possible happenings, but that is a much more complex and time-consuming way of getting the same answer.

Hint: The probability of getting at least 1 head is the same as:
( 1 - probability of never getting a head ).

this is equal to 1 minus the probability of getting tail in all trial

= 1 - (0.5^10)

= 0.999023438

h+t (10)

10/2=5
10/5=1/2

1/2=50% chance.

p(1)= 1 - 10/1024= 99.0234375