\displaystyle \textbf{Question 1: }\text{Suppose you drop a tie at random on the rectangular region shown in}
\displaystyle \text{figure. What is the probability that it will land inside the circle with diameter }1\text{ m?} \displaystyle \text{Answer:}
\displaystyle \text{Area of the rectangular region}=3\times2=6\text{ m}^2.
\displaystyle \text{Diameter of the circle}=1\text{ m}.
\displaystyle \therefore \text{Radius of the circle}=\frac{1}{2}\text{ m}.
\displaystyle \text{Area of the circle}=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}\text{ m}^2.
\displaystyle P(\text{tie lands inside the circle})=\frac{\text{Area of the circle}}{\text{Area of the rectangular region}}.
\displaystyle =\frac{\frac{\pi}{4}}{6}=\frac{\pi}{24}.
\displaystyle \therefore \text{The probability that the tie lands inside the circle is }\frac{\pi}{24}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In the accompanying diagram a fair spinner is placed at the centre }O\text{ of the}
\displaystyle \text{circle. Diameter }AOB\text{ and radius }OC\text{ divide the circle into three regions labelled }X,Y
\displaystyle \text{and }Z.\text{ If }\angle BOC=45^\circ,\text{ what is the probability that the spinner will land in region }X\text{?}
\displaystyle \text{(See figure).} \displaystyle \text{Answer:}
\displaystyle \text{Since }AOB\text{ is a diameter, }\angle AOB=180^\circ.
\displaystyle \angle BOC=45^\circ.
\displaystyle \therefore \text{Angle corresponding to region }X=180^\circ-45^\circ=135^\circ.
\displaystyle P(\text{spinner lands in region }X)=\frac{\text{Angle of sector }X}{360^\circ}.
\displaystyle =\frac{135^\circ}{360^\circ}=\frac{3}{8}.
\displaystyle \therefore \text{The probability that the spinner will land in region }X\text{ is }\frac{3}{8}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{A target shown in figure, }\text{ consists of three concentric circles of radii }3,7
\displaystyle \text{and }9\text{ cm respectively. A dart is thrown and lands on the target. What is the probability that}
\displaystyle \text{the dart will land on the shaded region?} \displaystyle \text{Answer:}
\displaystyle \text{The shaded region lies between the circles of radii }3\text{ cm and }7\text{ cm.}
\displaystyle \text{Area of the shaded region}=\pi(7^2-3^2).
\displaystyle =\pi(49-9)=40\pi\text{ cm}^2.
\displaystyle \text{Area of the entire target}=\pi(9)^2=81\pi\text{ cm}^2.
\displaystyle P(\text{dart lands on the shaded region})=\frac{\text{Area of the shaded region}}{\text{Area of the entire target}}.
\displaystyle =\frac{40\pi}{81\pi}=\frac{40}{81}.
\displaystyle \therefore \text{The probability that the dart lands on the shaded region is }\frac{40}{81}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In figure,}\text{ points }A,B,C\text{ and }D\text{ are the centres of four circles that}
\displaystyle \text{each have a radius of length one unit. If a point is selected at random from the interior}
\displaystyle \text{of square }ABCD,\text{ what is the probability that the point will be chosen from the shaded region?} \displaystyle \text{Answer:}
\displaystyle \text{Since each circle has radius }1\text{ unit and adjacent circles touch each other,}
\displaystyle AB=BC=CD=DA=2\text{ units}.
\displaystyle \therefore \text{Area of square }ABCD=2^2=4\text{ square units}.
\displaystyle \text{The four unshaded portions inside the square are four quarter-circles of radius }1\text{ unit.}
\displaystyle \therefore \text{Area of the four quarter-circles}=\pi(1)^2=\pi\text{ square units}.
\displaystyle \therefore \text{Area of the shaded region}=4-\pi\text{ square units}.
\displaystyle P(\text{point lies in shaded region})=\frac{\text{Area of shaded region}}{\text{Area of square }ABCD}.
\displaystyle =\frac{4-\pi}{4}=1-\frac{\pi}{4}.
\displaystyle \therefore \text{The required probability is }\frac{4-\pi}{4}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In figure, }\ JKLM\text{ is a square with sides of length }6\text{ units. Points }A
\displaystyle \text{and }B\text{ are the mid-points of sides }KL\text{ and }LM\text{ respectively. If a point is selected at}
\displaystyle \text{random from the interior of the square, what is the probability that the point will be chosen}
\displaystyle \text{from the interior of }\triangle JAB\text{?} \displaystyle \text{Answer:}
\displaystyle \text{Side of square }JKLM=6\text{ units.}
\displaystyle \therefore \text{Area of square }JKLM=6^2=36\text{ square units}.
\displaystyle \text{Since }A\text{ and }B\text{ are the mid-points of }KL\text{ and }LM,
\displaystyle KA=AL=LB=BM=3\text{ units}.
\displaystyle \text{Area of }\triangle JKA=\frac{1}{2}\times6\times3=9\text{ square units}.
\displaystyle \text{Area of }\triangle ALB=\frac{1}{2}\times3\times3=\frac{9}{2}\text{ square units}.
\displaystyle \text{Area of }\triangle JMB=\frac{1}{2}\times6\times3=9\text{ square units}.
\displaystyle \therefore \text{Area of }\triangle JAB=36-\left(9+\frac{9}{2}+9\right).
\displaystyle =36-\frac{45}{2}=\frac{27}{2}\text{ square units}.
\displaystyle P(\text{point lies inside }\triangle JAB)=\frac{\text{Area of }\triangle JAB}{\text{Area of square }JKLM}.
\displaystyle =\frac{\frac{27}{2}}{36}=\frac{27}{72}=\frac{3}{8}.
\displaystyle \therefore \text{The required probability is }\frac{3}{8}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In figure, }\text{ a square dart board is shown. The length of a side of the}
\displaystyle \text{larger square is }1.5\text{ times the length of a side of the smaller square. If a dart is thrown}
\displaystyle \text{and lands on the larger square, what is the probability that it will land in the interior of the}
\displaystyle \text{smaller square?} \displaystyle \text{Answer:}
\displaystyle \text{Let the side of the smaller square}=x\text{ units}.
\displaystyle \therefore \text{Side of the larger square}=1.5x=\frac{3x}{2}\text{ units}.
\displaystyle \text{Area of the smaller square}=x^2\text{ square units}.
\displaystyle \text{Area of the larger square}=\left(\frac{3x}{2}\right)^2=\frac{9x^2}{4}\text{ square units}.
\displaystyle P(\text{dart lands inside the smaller square})=\frac{\text{Area of smaller square}}{\text{Area of larger square}}.
\displaystyle =\frac{x^2}{\frac{9x^2}{4}}=\frac{4}{9}.
\displaystyle \therefore \text{The required probability is }\frac{4}{9}.
\displaystyle \\

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