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 $={1,2,3}$
$P(A)=$ Probability that number is multiple of 4
$P(B)=$ Probability that number is multiple of 6
$P(C)=$ Probability that number is multiple of 7
Now,
$P(A)=\frac{12}{50}, P(B)=\frac{8}{50}, P(C)=\frac{7}{50}$
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
$$ \begin{aligned} & P(A \cup B \cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0 \\ & \quad=\frac{21}{50} \end{aligned} $$
