Suppose n (n>1) points are placed at random on a circumference of a circle.
If P(n) denotes the probability that all n points lie on the same side of some diameter - find P(2), P(3) and P(4).
Draw the first point anywhere on the circle and draw a diameter through it. Draw the second point anywhere on the circle. Half the circle containing this second point determines the side on which all the rest of the points need to be. Therefore, clearly P(2)=1.
P(3) will clearly be the probability of the third point to be placed on the required half circle determined by the first 2 points. This prbability is P(3)=0.5, as there are only 2 equally likely half circles for the third point to be placed in. Likewise is the probability of each additional point to fall inside the required half circle equal to 0.5, but P(4) is the probability of both the third as well as the 4th point to fall as required, therefore P(4)=0.5*0.5 = 0.25
Solution
