The following continuous situations involve functions and co

The following continuous situations involve functions and conic functions. For each assume that you pick real numbers a and b at random between o and 10. Find the probability of the following.

3. The circle (x-a)^2+(y-b)^2 =1 lies entirely within a square with opposite vertices (0,0) and (10,10)

4. The ellipse x ² / a ² + y² / b ² = 1 has a vertical major axis

5. The hyperbola   x ² / a ² - y² / b ² = 1 has a horizontal transverse axis

6. the Interval from 4 to 6 is a part of the domain of the function y = x-a + b

Solution

3. The circle (x –a)² + (y –b)² = 1 has its center at (a,b) and its radius is 1. We presume that if the circle touches the square from within, then the circle is entirely within the square. For meeting this requirement, the distance of the center of the circle from the lines x = 0, x = 10 y = 0 and y = 10 has to be greater than or equal to 1. Thus, 1 a 9 and 1 b 9. Again we presume that a and b are integers ( as there are infinite number of real numbers between 0 and 10 and if a and b are not integers, the required probability cannot be computed). Then the probability of a and b fulfilling the conditions 1 a 9 and   1 b 9 is 9/11 each ( there are 11 integers between 0 and 10 and 9 integers between 1 and 9). Further, we know that P( A and B) = P(A)* P(B/A) where P(B/A) denotes” the probability of B, once A has happened”. Therefore, the probability of the required condition being met is 9/11*9/11 = 81/121 = 0.669 ( on rounding off to 3 decimal places).

4. The ellipse x ² / a ² + y² / b ² = 1 will have a vertical major axis if b > a. Now if a = 0, then b can be 1,2,3,…,10; if a =1, then b can be 2,3,4,…,10 and so on. Therefore, the required probability is 10/11*9/11*8/11* 7/11*6/11*5/11*4/11*3/11 *2/11*1/11 ( having chosen a, the numerators are the remaining choices for b; a can be any of the 11 integers ) = (10!/11¹⁰) = 3628800/ 25937424601 = 0.00014 ( on rounding off to 5 decimal places)

5. The hyperbola   x ² / a ² - y² / b ² = 1 with center at (0,0) will always have a horizontal transverse axis. Therefore, the probability of the hyperbola   x ² / a ² - y² / b ² = 1 having a horizontal transverse axis is 1.

6. The function y = (x-a) + b is an arm of a parabola and its domain is { x: x R and x a}.If the interval from 4 to 6 is a part of the domain of the function y = x-a + b, then a 4. Thus a can be 0,1,2,3 or 4. Therefore, the probability of the interval from 4 to 6 being a part of the domain of the function y = x-a + b is 5/11 = 0.4545… = 0.45 ( on rounding off to 2 decimal places)