Question

There is a circle with inner radius of 4 cm and outer radius of 6 cm. A coin is tossed onto this circle. What is the probability of the coin falling outside the smaller circle but inside the bigger circle?

• A. $\frac{2}{3}$
• B. $\frac{4}{9}$
• C. $\frac{5}{9}$
• D. $\frac{7}{9}$

Answer 1

The correct option is C $\frac{5}{9}$

The favourable outcome here is the coin falling in an area between the inner circle(of radius 4 cm) and the outer circle (of radius 6 cm).

The area of this required area $=\pi \times (6^2-4^2)$ $c{m}^2$

The total area where the coin is equally likely to fall $=\pi \times \left(6^2\right)$ $c{m}^2$

Therefore, the required probability $=\frac{\pi \times (6^2-4^2)}{\pi \times 6^2}$
$=\frac{36-16}{36}$
$=\frac{20}{36}$
$=\frac{5}{9}$