Consider a series of independent coin shots with the possibility of $p$ coming head in any shot. Let $Y_n$ be the numbers of heads in first $n$ consecutive shots of the coin. In this case find the probability of $\left \{ Y_{5}=3,Y_{8}=5,Y_{14}=9 \right \}$.
$\left \{ Y_{n},n=1,2,3,... \right \}$ is a Binomial process.
$Y_{8}-Y_{5}=Z_{6}+Z_{7}+Z_{8}=2$
$Y_{14}-Y_{8}=Z_{9}+Z_{10}+Z_{11}+...+Z_{14}=4$ then from here
$Y_{14}-Y_{5}=6$
$\Rightarrow P(\left \{ Y_{14}-Y_{5}=6\right \})=P(Y_{9}=6)=\binom{9}{6}p^{6}q^{3}$
But the right answer is $450p^{9}q^{5}$
Where did I go wrong? Any help will be appreciated.
Hint:
$$Y_5=3\text{ and }Y_8=5\text{ and }Y_{14}=9\iff Y_5=3\text{ and }Y_8-Y_5=2\text{ and }Y_{14}-Y_8=4$$
So you should go for finding: $$P(Y_5=3,Y_8-Y_5=2,Y_{14}-Y_8=4)$$ where the rv's $Y_5$ ,$Y_8-Y_5$, $Y_{14}-Y_8$ are independent.