Explanation

(a)
Given, $\sigma_{x}^{2} = 9$
And lines of regression are
$4x - 5y + 33 = 0,20x - 9y - 10 = 0$
Ie, $y = \frac{4}{5}x + \frac{33}{5}and\ x = \frac{9}{20}y + \frac{10}{20}$
$\therefore$Regression coefficient are
$b_{yx} = \frac{4}{5}and\ b_{xy} = \frac{9}{20}$
Now, $b_{yx} = \frac{cov(x,y)}{\sigma_{x}^{2}}$
$\Rightarrow cov(x,y) = \frac{4}{5} \times 9 = \frac{36}{5}$
And $b_{xy} = \frac{cov(x,y)}{\sigma_{y}^{2}}$
$\Rightarrow \sigma_{y}^{2} = \frac{36}{5} \times \frac{20}{9} = 16$
Now, $p(x,y) = \frac{cov(x,y)}{\sigma_{x}\sigma_{y}} = \frac{36}{5 \times 3 \times 4} = 0.6$