An integer is chosen at random from the integers 1, 2, 3, . . . . ....

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 $P\left(4\right) = \dfrac{12}{50}$

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

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

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

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

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

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

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

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