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

<p>Given set $=\{1,2,3, \ldots \ldots . .50\}$</p> <p>$\mathrm{P}(\mathrm{A})=$ Probability that number is multiple of 4</p> <p>$\mathrm{P(B)}=$ Probability that number is multiple of 6</p> <p>$\mathrm{P}(\mathrm{C})=$ Probability that number is multiple of 7</p> <p>Now,</p> <p>$$\mathrm{P}(\mathrm{A})=\frac{12}{50}, \mathrm{P}(\mathrm{B})=\frac{8}{50}, \mathrm{P}(\mathrm{C})=\frac{7}{50}$$</p> <p>again</p> <p>$$\begin{aligned} & \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{50}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{50}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{50} \\ & \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0 \end{aligned}$$</p> <p>Thus</p> <p>$$\begin{aligned} P(A & \cup B \cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0 \\ & =\frac{21}{50} \end{aligned}$$</p>