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put this solution on YOUR website!CIRCUMFERENCE OF OUTER CIRCLE =16*PI
DIAMETER=16*PI/PI=16...........RADIUS =8
AREA=8*8*PI=64*PI
INSCRIBED CIRCLE WILL HAVE ITS RADIUS EQUAL TO ALTITUDE OF ANY EQUILATERAL TRIANGLE FORMED BY ONE SIDE OF HEXAGON WITH CENTRE OF BIG CIRCLE.
SIDE OF HEXAGON = RADIUS OF CIRCUM CIRCLE =8
ALTITUDE = 8 *SIN60=8*SQRT.(3)/2=4SQRT.3
SO AREA OF INNER CIRCLE=PI*4*4*(SQRT.3)^2=48*PI
SO PROBBILITY OF A RANDOM POINT SELECTED IN BIG CIRCLE WILL FALL IN THE SMALLER CIRCLE=AREA OF SMALL CIRCLE/AREA OF BIG CIRCLE= 48*PI/64*PI=3/4=0.75