<s>[INST] Let $X$ denote the number of tosses required to get the 5th head and $Y$ the number between the 6th and 7th heads. Are $X$ and $Y$ independent? 
Let $X$ denote the number of tosses required to get the 5th head and $Y$ the number between the 6th and 7th heads. Are $X$ and $Y$ independent?

Y will always depend on X . NO ?
i know geometric distribution has lack of memory property . but in this the underlying distribution is i think negative binomial. 
i think 'no, they are not independent ' . help please. 
 [/INST] What you want to show is that 
$$
P(X=x \wedge Y=y) = P(X=x)P(Y=y)
$$
for all values $x$ and $y$ of $X$ and $Y$.
Let $Z$ be the number of tosses between the 5th and 6th head.  Then
\begin{align*}
    P(X=x \wedge Y=y)
    &= \sum_{z=1}^\infty \underbrace{\binom{x-1}{4}p^5q^{x-5}}_{X=x} \underbrace{\vphantom{\binom{x-1}{4}}q^{z-1}p}_{Z=z} \underbrace{\vphantom{\binom{x-1}{4}}q^{y-1}p}_{Y=y} \\
    &= \binom{x-1}{4}p^5q^{x-5} \cdot q^{y-1}p \cdot \sum_{z=1}^\infty q^{z-1}p \\
    &= \binom{x-1}{4}p^5q^{x-5} \cdot q^{y-1}p \cdot \frac{p}{1-q} \\
    &= \binom{x-1}{4}p^5q^{x-5} \cdot q^{y-1}p = P(X=x)P(Y=y)
\end{align*}
Once you separate the events like this it boils down to the tosses being independent of each other.
</s>