How to find probability for this situation? You flip a fair coin $5$ times. What is the probability that the first flip results in heads or the fifth flip results in heads?
(a) $\frac14$ 
(b) $\frac12$ 
(c) $\frac34$
(d) $1$
the ans is $\frac34$ but i dont understand how .
i figured that the sample space will have $32$ elements ($2^5$)
and im guessing once we get the probability for the first flip= heads, we will just need to multiply it by $2$ as last flip=heads . or is this approach wrong?
 A: *

*Since it asks only the probability that first and fifth coins land head, it means that other coins can be anything. 

*So there are 4 possible ways in which 1st and 5th coins can land. And there are 8 ways in which middle 3 coins can land.

*Of the 4 possible ways that 1st and 5th coin can land 3 are favorable, i.e HH HT TH. For each of these possibilities the middle three coins can land in 8 ways. Making total favorable outcomes = $3\times 8 = 24$.

*Hence probability is $24/32=3/4$

A: Hint:
Use this formula:
$$P(A \cup B) = P(A)+P(B)-P(A \cap B)$$
A natural choice of $A$ is $\{ \text{first flip results in head }\}$
A: As it says ''or'' we have to consider all three situations;
i)First flip=Head, Fifth flip=Tail
i)First flip=Tail, Fifth flip=Head
i)First flip=Head, Fifth flip=Head 
So there lies the situations we have, what we can do to get things shorter is to subtract the situations in which the first and fifth flips are tails. The entire set of options is $2^5$ because we have $2$ situations each time we flip the coin so $s(E)=2^5$ and now I will consider $1\cdot2\cdot2\cdot2\cdot1=2^3$ as the situations in which the first and fifth flip are tails,(I didn't multiply them with a factorial because I know the order of them); overall: $$P(A')=\frac{s(A)}{s(E)}=\frac{2^3}{2^5}=\frac{1}{4}$$  $$P(A)=1-P(A')=1-\frac{1}{4}=\frac{3}{4}$$ which is the desired answer. Normally it is not that slow to apply this method. I've slowed things down a little to be more pedantic.
