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Question: A point is selected at random from the interior of the circle. The probability that the point is clo...

A point is selected at random from the interior of the circle. The probability that the point is closer at the centre than the boundary of the circle is:
A. 34\dfrac{3}{4}
B. 12\dfrac{1}{2}
C. 14\dfrac{1}{4}
D. none of these

Explanation

Solution

let us assume the circle of the radius rr so now if the point lie within the radius r2\dfrac{r}{2} then it will be close to the circle then the probability of the point to be closer to the centre will be the ratio of the area of the circle with the radius r2\dfrac{r}{2} and the area of the circle with the radius rr.

Complete step by step solution:
Here we are given in the question that we have to select the point from the interior of the circle such that it is closer to the centre. So let us assume the circle of the radius rr and the centre O.

So we need to consider the point such that it is closer to the point O so we draw another concentric circle of the radius r2\dfrac{r}{2} so that when it lies within the circle of the radius r2\dfrac{r}{2} then it will be considered
As closest to the centre of the radiusrr.
If it is outside the circle of the radius r2\dfrac{r}{2} and inside the circle of the radius r then it will be considered closer to the circumference.
We also know thatprobability=number of favourable outcometotal number of outcomes{\text{probability}} = \dfrac{{{\text{number of favourable outcome}}}}{{{\text{total number of outcomes}}}}
So the number of possible outcomes will be the area of the circle of the radius r2\dfrac{r}{2} in which the point lies closer to the centre so the possible area=π(r2)2 = \pi {\left( {\dfrac{r}{2}} \right)^2}
Total area would be πr2\pi {r^2}
So probability=π(r2)2πr2=r24r2=14{\text{probability}} = \dfrac{{\pi {{\left( {\dfrac{r}{2}} \right)}^2}}}{{\pi {r^2}}} = \dfrac{{{r^2}}}{{4{r^2}}} = \dfrac{1}{4}

Hence option C is correct.

Note:
We should know that the probability of any two events A and B is given by
P(AB)=P(AB)P(B)P\left( {\dfrac{A}{B}} \right) = \dfrac{{P(A \cap B)}}{{P(B)}} and here P(AB)P\left( {\dfrac{A}{B}} \right) represents the probability of the occurrence of A if B has already occurred and P(B)P(B) is the probability of the occurrence of event BB