You have a circle with radius r. What is the probability that a point lies more towards the circumference than towards the centre.

Solution
Probability of point lying close to circumference would be mean point is in the circle with area from r/2 from center till outside in the figure below it would be in blue part of the circle.

Probability = Area covered by circle from point r/2 towards circumference/ Area of the circle
In the above given circle formula would be = Area of blue circle/ Area of circle
= pi r pow 2/ 4 / pi r 2 = 1/4

minimize the time taken by all 4 to cross the bridge.

There is a bridge and to travel on it there should be a torch light available.
Only two persons can cross the bride at a time. There are 4 persons on the other side of the bridge. They require 1,2,5,10 minutes respectively to cross the bridge. How much time should it take for all 4 to go to the other side of the bridge.

Remember there is only one torch light. So if two persons cross the bridge with torch light, one has to come back with torch light for others to use torch light. If person with 1 min and 10 minutes cross the bridge together, they take 10 minutes together to cross the bridge(maximum of the time taken by two people).

Your criteria is to minimize the time taken by all 4 to cross the bridge.

How many regions (open or closed) can n straight lines give at the maximum?

Let f(n) denote the number of regions for n lines no three of which coincide in one point. This latter condition will guarantee the maximum. Obviously, f(0)=1, f(1)=2, f(2)=4, f(3)=7, etc.

Now, given n-1 lines, there are already f(n-1) regions. With the introduction of the n-th line, imagine the points of intersection of the pre-existing n-1 lines fall on one side of the n-th line. So, on one side of the n-th line -- where the points of intersection lie -- you still have f(n-1) regions, whereas on the other side of the n-th line, you now have n regions. So, for these n lines before us, the total number f(n) of regions must be equal to f(n-1)+n.

So, f(n)=f(n-1)+n. With the initial condition(s) given earlier, this gives us f(n)=1+n(n+1)/2.

find the optimal start point on the track such that you never run out of fuel and complete circuit.

You have a circular track containing fuel pits at irregular intervals. The total amount of fuel available from all the pits together is just sufficient to travel round the track and finish where you started. Given the the circuit perimeter, list of each fuel pit location and the amount of fuel they contain, find the optimal start point on the track such that you never run out of fuel and complete circuit.