Question 394210
It is equally likely that a fair coin will come up H or T
In probabilities we say
P(H) = 1/2
P(T) = 1/2
Note that P(H) + P(T) must add up to 1, since 1 = certainty, and
it is certain that nothing else can happen (no losing coin allowed)
----------------------------
P(0 tails) = P(5 heads) for 5 tosses
What you're doing is tossing the coin and saying I want Heads 5 time
in a row, or H and H and H and H and H. When ever you want to find
P(A and B and C...) you must MULTIPLY the separate probabilities, so
1st toss P(H) = 1/2
2nd toss P(H) = 1/2
3rd toss P(H) = 1/2
4th toss P(H) = 1/2
5th toss P(H) = 1/2
P(5 heads) = (1/2)^5
(1/2)^5 = 1/32 answer
--------------------
Exactly 1 tail means you only want to see
either:
THHHH
HTHHH
HHTHH
HHHTH
HHHHT
In the previous problem, I found there are 32 possible outcomes
for 5 tosses , 1 of which was HHHHH. Now I want only 1 of the
above 5 choices.
So, P(exactly 1 T) = 5/32
Note that you add the P of each because you want P (THHHH OR HTHHH OR ... etc.)
and generally P(A or B) = P(A) + P(B)
----------------------
At most 2 tails
This means  1 tail is OK OR 2 tails is OK , but all others are rejected
I want to find P(1 tail) + P(2 tails)
1 tail would be
THHHH or 
HTHHH or
HHTHH or
HHHTH or
HHHHT 
So, P(1 tail) = 5/32
What are all the possible ways you can get exactly 2 tails and 3 heads?
That is the permutation of 5 things taken 2 at a time.
You can look that up. I'm out of time, sorry