Probability Question 6
Question 6 - 2024 (29 Jan Shift 2)
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
(1) $\frac{8}{25}$
(2) $\frac{21}{50}$
(3) $\frac{9}{50}$
(4) $\frac{14}{25}$
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Answer (2)
Solution
Given set $={1,2,3$
$\mathrm{P}(\mathrm{A})=$ 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
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{50}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{50}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{50}$
$\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0$
Thus
$$ \begin{aligned} & P(A \cup B \cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0 \ & \quad=\frac{21}{50} \end{aligned} $$
