Two events X and Y are such that $P(X) = \frac{1}{3}, P(Y)= n$ and the probability of occurrence of at least one event is 0.8. If the events are independent, then the value of n is:

$\frac{7}{10}$

The correct answer is Option (3) → $\frac{7}{10}$

We are given two independent sets X and Y with probabilities,

$P(X)=\frac{1}{3},P(Y)=n$

$P(X∪Y)=P(X)+P(Y)-P(X)P(Y)$

$0.8=\frac{1}{3}+n-\left(\frac{1}{3}×n\right)$

$⇒2.4=1+3n-n$

$⇒1.4=2n$

$⇒n=0.7=\frac{7}{10}$