Let $A=\left\{1,2,3,4,5,.........,50\right\}$

Now, the probability of choosing a number which is a multiple of $4$ will be $P\left(4\right)=\frac{12}{50}$

Similarly, the probability of choosing a number which is a multiple of $6$ will be $P\left(6\right)=\frac{8}{50}$

And the probability of choosing a number which is a multiple of $7$ will be $P\left(7\right)=\frac{7}{50}$

Also, the probability of choosing a number which is a multiple of $4 \& 6$ will be $P\left(4\cap 6\right)=\frac{4}{50}$

And the probability of choosing a number which is a multiple of $6 \& 7$ will be $P\left(6\cap 7\right)=\frac{1}{50}$

And the probability of choosing a number which is a multiple of $4 \& 7$ will be $P\left(4\cap 7\right)=\frac{1}{50}$

And finally, the probability of choosing a number which is a multiple of $4, 6 \& 7$ will be $P\left(4\cap 6\cap 7\right)=0$

Hence, $P\left(4\cup 6\cup 7\right)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0$

$\Rightarrow P\left(4\cup 6\cup 7\right)=\frac{21}{50}$