Here, we define event
A = Getting a prime number on die.
B = Getting an odd number on die.
Given that number appearing on the die is odd, so we need to calculate probability that number is prime too.
Prime number on die = 2, 3, 5
$\mathit{P}\left(\mathit{A}\right)=\frac{3}{6}=\frac{1}{2}$
Odd number on die $=1,3,5$
$\mathit{P}\left(\mathit{B}\right)=\frac{3}{6}=\frac{1}{2}$
$\mathit{A}\cap \mathit{B}=\left\{3,5\right\}$
$\mathit{P}\left(\mathit{A}\cap \mathit{B}\right)=\frac{2}{6}=\frac{1}{3}$
$\mathit{P}\left(\mathit{A}\mid \mathit{B}\right)=\frac{\mathit{P}\left(\mathit{A}\cap \mathit{B}\right)}{\mathit{P}\left(\mathit{B}\right)}=1∕31∕2=\frac{\frac{2}{3}}{.}$
So assertion is true.
In reason, given that $\mathit{P}\left(\mathit{A}\mid \mathit{B}\right)=\frac{\mathit{P}\left(\mathit{A}\cup \mathit{B}\right)}{\mathit{P}\left(\mathit{B}\right)}$, which is wrong statement, as
$\mathit{P}\left(\mathit{A}\mid \mathit{B}\right)=\frac{\mathit{P}\left(\mathit{A}\cap \mathit{B}\right)}{\mathit{P}\left(\mathit{B}\right)}$