JEE Main 2024 — 29 January, Shift 2 — Question 19

Given set $=\{1,2,3$,….,$50\}$

$\mathrm{P}(\mathrm{A})=$ Probability that number is multiple of 4 = $\frac{12}{50}$

$P(B)=$ Probability that number is multiple of 6 = $\frac{8}{50}$

$\mathrm{P}(\mathrm{C})=$Probability that number is multiple of 7 = $\frac{7}{50}$

$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{50}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{50}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{50}$, $\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0$

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

$\boxed{\frac{21}{50}}$