Given that [tex]P(Q)=\dfrac{3}{5},P(R)=\dfrac{1}{3}[/tex]
Also,
[tex]P(Q\wedge R)=P(Q)\cdot P(R)=\dfrac{3}{5}\cdot\dfrac{1}{3}=\dfrac{1}{5}[/tex]
We can conclude that,
[tex]P(Q\vee R)=P(Q)+P(R)=\dfrac{3}{5}+\dfrac{1}{3}=\boxed{\dfrac{14}{15}}[/tex]
The answer is B.
Hope this helps.
r3t40
