An integer is chosen at random from the integers 1, 2, 3, …. 50. Twice  the probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is

$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-(P\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)$

$P\left(A\right)=\frac{12}{50};P\left(B\right)=\frac{8}{50};P\left(C\right)=\frac{7}{50}$

$P(A\cap B)\text{}=\frac{2}{50};P(B\cap C)=\frac{1}{50}$

$$\begin{gathered} P(C\cap A)=\frac{1}{50},P(A\cap B\cap C)=0 \\ P(A\cup B\cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}=\frac{21}{50} \end{gathered}$$