In binomial distribution with $n = 10 $ and $P=\frac{1}{3}$, the probability of the event that unequal number of failures and successes occur is :

$\frac{5665}{6561}$

The correct answer is Option (1) → $\frac{5665}{6561}$

For Binomal distribution,

$P(X=k)=\left({^nC}_k\right)P^kq^{n-k}$

$∴P(X=5)={^{10}C}_5(\frac{1}{3})^5(\frac{2}{3})^5$

$=182$

$P(X=5)=182×(\frac{1}{3^5})×(\frac{2}{3})^5$

$=\frac{182×32}{3^{10}}=\frac{5324}{59049}$

$P(X≠5)=1-P(X=5)=1-\frac{5324}{59049}$

$=\frac{5665}{6561}$