Let X be a random variable which assumes value $x_1, x_2, x_3, x_4$ such that $2P(x=x_1)=3P(x=x_2)=P(x=x_3)= 5P(x=x_4), $ then Probability $P(x=x_3) $ will be

$\frac{15}{61}$ $\frac{30}{61}$ $\frac{6}{61}$ $\frac{45}{61}$

$\frac{30}{61}$