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A number $\mathrm{x}$ is chosen at random from the set $\{1$, $2,3,4, \ldots ., 100\}$. Define the event: $\mathrm{A}=$ the chosen number $x$ satisfies $\frac{(x-10)(x-50)}{(x-30)} \geq 0$
Then $\mathrm{P}(\mathrm{A})$ is:
Options:
Then $\mathrm{P}(\mathrm{A})$ is:
Solution:
2173 Upvotes
Verified Answer
The correct answer is:
$0.71$
$0.71$
Given $\frac{(x-10)(x-50)}{(x-30)} \geq 0$
Let $x \geq 10, x \geq 50$ equation will be true $\forall x \geq 50$ as $\left(\frac{x-50}{x-30}\right) \geq 0, \forall x \in[10,30)$
$$
\frac{(x-10)(x-50)}{x-30} \geq 0 \quad \forall x \in[10,30)
$$
Total value of $x$ between 10 to 30 is 20 . Total values of $x$ between 50 to 100 including 50 and 100 is 51 .
Total values of $x=51+20=71$
$$
P(\mathrm{~A})=\frac{71}{100}=0.71
$$
Let $x \geq 10, x \geq 50$ equation will be true $\forall x \geq 50$ as $\left(\frac{x-50}{x-30}\right) \geq 0, \forall x \in[10,30)$
$$
\frac{(x-10)(x-50)}{x-30} \geq 0 \quad \forall x \in[10,30)
$$
Total value of $x$ between 10 to 30 is 20 . Total values of $x$ between 50 to 100 including 50 and 100 is 51 .
Total values of $x=51+20=71$
$$
P(\mathrm{~A})=\frac{71}{100}=0.71
$$
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