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An integer is chosen at random from the integers $\{1,2,3, \ldots \ldots . .50\}$. The probability that the chosen integer is a multiple of atleast one of $4,6$ and $7$ is
$\frac{8}{25}$
$\frac{21}{50}$
$\frac{9}{50}$
$\frac{14}{25}$
Solution
Given set $=\{1,2,3, \ldots \ldots . .50\}$
$\mathrm{P}(\mathrm{A})=$ Probability that number is multiple of $4$
$P(B)=$ Probability that number is multiple of $6$
$\mathrm{P}(\mathrm{C})=$ Probability that number is multiple of $7$
Now,
$\mathrm{P}(\mathrm{A})=\frac{12}{50}, \mathrm{P}(\mathrm{B})=\frac{8}{50}, \mathrm{P}(\mathrm{C})=\frac{7}{50}$
again
$ P(A \cap B)=\frac{4}{50}, P(B \cap C)=\frac{1}{50}, P(A \cap C)=\frac{1}{50} $
$ P(A \cap B \cap C)=0$
Thus
$ 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}$
