probability of obtaining an even number of heads in $5$ tosses, zero being treated as an even number
number of event = $0$ head or $2$ head or $4$ head
Probability of head =$ \dfrac{2}{3}$
Probability of tail = $\dfrac{1}{3}$
Probability = ${^5}C{_0}\left(\dfrac{2}{3}\right)^{0}\left(\dfrac{1}{3}\right)^{5}+{^5}C{_2}\left(\dfrac{2}{3}\right)^{2}\left(\dfrac{1}{3}\right)^{3} +{^5}C{_4}\left(\dfrac{2}{3}\right)^{4}\left(\dfrac{1}{3}\right)^{1}$
=$\dfrac{121}{243}.$
Option A