There are some experiments in which the outcomes cannot be identified discretely. For example, an ellipse of eccentricity $$22/3$$ is inscribed in a circle and a point within the circle is chosen at random. Now, we want to find the probability that this point lies outside the ellipse. Then, the point must lie in the shaded region shown in Figure. Let the radius of the circle be a and length of minor axis of the ellipse be $$2b$$. Given $$that1$$−$$b2a2=89$$ or $$b2a2=19Then$$, the area of circle serves as sample space and area of the shaded region represents the area for favorable cases. Then, required probability $$isp=$$ Area of shaded region Area of circle $$=$$π$$a2$$−πab$$π$$$$a2=1$$−$$ba=1$$−$$13=23Now$$ answer the following questions.A point is selected at random inside a circle. The probability that the point is closer to the center of the circle than to its circumference isTwo persons A and B agree to meet at a place between $$5$$ and $$6$$ pm. The first one to arrive waits for $$20$$ min and then leave. If the time of their arrival be independent and at random, then the probability that A and B meet is

Question

There are some experiments in which the outcomes cannot be identified discretely. For example, an ellipse of eccentricity $$22/3$$ is inscribed in a circle and a point within the circle is chosen at random. Now, we want to find the probability that this point lies outside the ellipse. Then, the point must lie in the shaded region shown in Figure. Let the radius of the circle be a and length of minor axis of the ellipse be $$2b$$. Given $$that1$$−$$b2a2=89$$ or $$b2a2=19Then$$, the area of circle serves as sample space and area of the shaded region represents the area for favorable cases. Then, required probability $$isp=$$ Area of shaded region  Area of circle $$=$$π$$a2$$−πab$$π$$$$a2=1$$−$$ba=1$$−$$13=23Now$$ answer the following questions.A point is selected at random inside a circle. The probability that the point is closer to the center of the circle than to its circumference isTwo persons A and B agree to meet at a place between $$5$$ and $$6$$ pm. The first one to arrive waits for $$20$$ min and then leave. If the time of their arrival be independent and at random, then the probability that A and B meet is

Infinity Learn · Accepted Answer

For the favorable cases, the points should lie inside the concentric circle of radius $$r/2$$. So the desired probability is given by Area of smaller circle  Area of larger circle $$=$$π$$r22$$$$π$$$$r2=14Let$$ A and B arrive at the place of their meeting 'a' minutes and 'b' minutes after $$5$$ pm. Their meeting is possible only if|a−b|≤$$20$$          (1)Clearly$$, $$0$$≤a≤$$60$$ and $$0$$≤b≤$$60$$. Therefore, a and b can be selected as an ordered pair $$(a$$, $$b) from the set [$$0$$, $$60$$] x [$$0$$, $$60$$].Alternatively, it is equivalent to select a point (a$$, $$b) from the square OPQR, where P is (60$$, $$0) and R is (0$$, $$60) in the Cartesian plane. Now,|a−b|≤$$20$$⇒−$$20$$≤a−b≤$$20Therefore$$, points (a$$, $$b) satisfy the equation −$$20$$≤x−y≤$$20$$.Hence, favorable condition is equivalent to selecting a point from the region bounded by y≤$$x+20$$ and y≥x−$$20$$. Therefore, the required probability is Area of OABQCDO Area of square $$OPQR=$$[Ar⁡(OPQR)−$$2Ar$$⁡$$($$Δ$$APB)$$]Ar⁡$$(OPQR)=60$$×$$60$$−$$22$$×$$40$$×$$4060$$×$$60=59$$

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