A coin is tossed 4 times The result is either (H) or  (T).
 Define the events


*

*D: ”The fourth toss is tails” and

*F:
”Heads and tails alternate”. 


Additionally,
determine: $$D ∩ F$$
Now for $P (D)$ there are 8 possibilities so then: $$P (D) = 1/(2^8) $$ but my friend said that it needs to be: $$8/(2^8) $$
Is that right? If so, why?
Edit:tried to make it more clear. I already know how to get P (F) and $$D ∩ F$$. My main confusion is just with P (D)
 A: $P(D)$ is $1/2$
Now let us see how...
By Laplace's law, Probability can be expressed as  the ratio of the number of favourable events to the total number of events.
Now if you know how to count permutations, you will see that there are clearly $2×2×2×1$ favourable events and $2×2×2×2$ total number of events. Taking the ratio, you get $1/2$    
Let us now find $P(F)$
For F, there are two favourable cases, namely, HTHT,THTH. And the total cases remain same.
So, $P(F)=1/2^3$   
Finally, $$P(D \cap F)=P(D)×P(F)=1/2^4$$
EDIT: Another way of finding $P(D)$
Clearly $P(D)=1/2$ because what we get on coins 1,2 and 3 has no effect on what we get on 4(independant events) and also there are no restrictions on the outcome of 1,2,3.
2nd EDIT  Let us define four events:.
1-Getting either a head or a tail on tossing coin 1.
2-Getting either a head or a tail on tossing coin 2.
3.Getting either a head or a tail on tossing coin 3.
4. Getting a tail on tossing coin 4.
NB All these events are independent. We are not tossing the other coins when we are tossing some $i$th coin.
Therefore, clearly $P(D)=1•1•1•1/2=1/2$
A: First, $P(D)=\frac12$, because other tosses do not matter.
For $P(F)$, we see that there are $2^4$ possible outcomes when tossing a coin 4 times. Of these, only 2 are good: THTH and HTHT. This makes a total of $P(F)=\frac{2}{2^4}=\frac{1}{2^3}=\frac{1}{8}$.
Now we combine $D$ and $F$. Then, only HTHT is left. So we have $P(D\cap F)=\frac{1}{2^4}=\frac{1}{16}$.
