Find the probability that a dart hits the region X(inner circle) if the radius of the outer circle is $x$ times radius of the inner circle (x>2).

$\frac{1}{x}$

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$\frac{1}{x^{2}}$

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$\frac{1}{x^{3}}$

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Solution

The correct option is B
$\frac{1}{x^{2}}$

Let the radius of innermost circle = y

$\Rightarrow$ radius of outermost circle = $y x$

Probability that a dart hits region X = $\frac{Area (innermost circle)}{Area (outermost circle)} = \frac{π . y^{2}}{π . y^{2} x^{2}} \Rightarrow \frac{1}{x^{2}}$

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Find the probability that a dart hits the region X(inner circle) if the radius of the outer circle is $x$ times the radius of the inner circle.

Q. If the radius of outer circle is x times radius of inner circle, then the probability that a dart hits the inner circle is

\frac{1}{x^{2}}

Find the probability that a dart hits the region X(inner circle) if the radius of the outer circle is $x$ times radius of the inner circle (x>2).

In a circular dartboard of radius 20 cm, there are 5 concentric circles. the radius of each inner concentric circle is 4 cm less than the outer concentric circle. Find the probability that a dart hits anywhere in the smallest circle assuming that the dart doesn't hit on the boundary of any circle.

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Find the probability that a dart hits the region X(inner circle) if the radius of the outer circle is x times radius of the inner circle (x>2).

The correct option is B 1x2 Let the radius of innermost circle = y ⇒radius of outermost circle = yx Probability that a dart hits region X = Area (innermost circle)Area (outermost circle)=π.y2π.y2x2⇒1x2

Find the probability that a dart hits the region X(inner circle) if the radius of the outer circle is $x$ times radius of the inner circle (x>2).

The correct option is B
$\frac{1}{x^{2}}$

Let the radius of innermost circle = y

$\Rightarrow$ radius of outermost circle = $y x$

Probability that a dart hits region X = $\frac{Area (innermost circle)}{Area (outermost circle)} = \frac{π . y^{2}}{π . y^{2} x^{2}} \Rightarrow \frac{1}{x^{2}}$