Solution to Probability: Catch the train

There are many ways to solve this problems including the ones using graphs or integral calculus. Although I've always got good grades for it, I hate advanced mathematics. At least I do not like it when I am at leisure, trying to solve really nice puzzles while stretching on a couch! So, the solution that I would offer would not include any calculus or graphs.

The answer to this puzzle is 7/8

The solution that I used is as follows. Let us find the probability of the man missing the train. Let us divide the total time interval in equal intervals of 5 minutes each:

09:55 to 10:00 T1

10:00 to 10:05 T2

10:05 to 10:10 T2

The man misses the train if he arrives within T2, and the train too leaves within T2, and the train leaves before the man arrives.

The probability of the man arriving within T2 is 1/2. Similarly, the probability of the train leaving within T2 is 1/2.

Now, we need the probability of the train leaving before the man arrives. Now, since the train's departure and the man's arrival are two independent events (with uniformly distributed probabilities), by symmetry it is equally probable that one event occurs before or after the other. Hence the probability of the train leaving before the man arrives (within T2) is 1/2.

So, the probability of the man missing the train is (1/2)*(1/2)*(1/2) = 1/8

The probability that he catches the train, hence, is 7/8.

Solution to the Centrifuge Problem

This is the solution to the problem discussed at http://ankit-mathur.blogspot.com/2013/06/physics-centrifuge-problem.html

We basically need to find out the system which has more kinetic energy.

Both in system 1 and system 2, the disk rotates at the same angular velocity. Now, since the coin is also rotating in the first case, it would have additional kinetic energy. The kinetic energy of the first system is equal to the the sum of the kinetic energy of the second system and the kinetic energy of the coin rotating on its own axis. Hence, the first system has more energy.

Hence we will require more electricity to start the first system from rest.

Water level problem

This is the solution to the water level problem discussed in my blog ankit-mathur.blogspot.in (alternatively ankitmathur.in).

The box is accelerating in the direction BA, creating a pressure differential between two ends of the box. Since the acceleration is constant, the ratio of pressure at the front and rear end of the box is constant, and so is the ratio between the heights in the tubes.

In simpler words, if the box is accelerating in direction BA then there is a pseudo force acting on the water in the direction AB, which is pushing water in B tube higher.

The 8 Ants Problem - The Time Dimension

This one's about the 8 ants problem i discussed in my earlier post. It has just grown more interesting! You had to prove that each ant meets every other ant at least once. A friend of mine just got another witty solution to it. Probably the 'elegant' solution that the friend who first put this puzzle to me talked about.

Consider a three dimensional space with time as the third dimension. The plane in which the ants are moving would be the x-y plane. Now, let us imagine, how the path of the ants looks like - many skew straight lines in the 3-D space. The z coordinate of these lines represents time. Now, if two of the ants meet anywhere at any point in time, their corresponding lines will intersect somewhere: somewhere in the 3-D space, and the z-coordinate of the intersection point would represent the time when the meeting occurred. Now, let lines A and B represent the ants which meet all other ants. These two lines together define a plane (statement 1). Any other line which intersects these two lines has to be coplanar with A and B (statement 2). It means all lines are coplanar. And since all coplanar lines intersect (unless they are parallel) (statement 3), we can say that all ants meet at some point of time (statement 4). Hence proved.

Give some thought to this before you read further.

Being unarguably a more interesting solution than the one I suggested, this one has some terrible flaws. If you can visualize quickly in 3-D, the first flaw would manifest itself quite clearly - what if one of these lines is not coplanar, but meets lines A and B at their point of intersection (i.e. the three lines intersect at the same point)? Now, this partcular line meets A and B and does not necessarily meet other lines. Hence, our solution breaks at this point. However, we can make minor alterations in the problem statement itself, to prevent thee ants from meeting at the same point, or something of that sort.

What really breaks this solution down is this: some of the lines intersect below the t=0 (or z=0) plane. So these meetings, in a sense, are virtual. Since these lines never met in t >= 0 space, the ants never actually met. Any meeting before t=0 cannot be considered as a real meeting.

Hence, the solution breaks down here. A nice thought, though!

The 8 ants problem

Here's the solution to the 8 ants problem:

Let all ants have different velocities. Since each ant is moving with some constant velocity, the relative velocity between any two ants (V1-V2) will always be constant. Constant velocity means straight line displacement. Hence,

(1) Relative displacement of any ant with respect to any other ant will be a straight ine.

Meaning, if ant A1 looks at ant A3, it will see A3 as moving in a straight line (relative to A1). Now, let A1 and A2 be those ants about which it is given that they meet all others. Let us see from the frame of reference of A1 (A1 is stationary at origin in this frame). A1 will see all ants moving at different velocities along straight lines. But since A1 meets all ants at least once,

(2) The straight line paths of all ants must pass through the origin.

Also, it is true that:
(3) Two straight lines can intersect only once.

Now we know that the locus of all ants are straight lines passing through the origin. Also, if any ant has to meet any other ant, it can do so only at the origin (by (3)). Now, since ant A2 meets all other ants, we can be sure that the meeting happens only at the origin. It is only possible if all the ants reach the origin at the same time. I means that each ant meets every other ant at origin. Hence Proved.

Now, there is a small issue. Not with the solution, but with the problem itself. It is not valid when all the ants are moving in one straight line. (because then, postulate 3 won't hold). I can give you one simple example to show that. Suppose A2 ... A8 are moving in the same direction in a straight line. V2>V3=V4=V5=V6=V7=V8. A2 is lagging all others at this point in time. And A1 is moving in the opposite direction towards all others. Hence, A1 will meet all ants. Also, A2 will meet all ants, as its velocity is greater than all others. But V3..V8 will never meet as they are moving with the same speed along the same line. Hence all ants do not meet.

The prisoners and the warden puzzle (survival instinct) solution

Following is the solution of one of the most intriguing of the puzzles I've ever come across, the Survival Instinct puzzle, which I discussed earlier in this blog.

The team nominates a leader. The group agrees upon the following rules:

The leader is the only person who will announce that everyone has visited the switch room. All the prisoners (except for the leader) will flip the first switch up at their very first opportunity, and again on the second opportunity. If the first switch is already up, or they have already flipped the first switch up two times, they will then flip the second switch. Only the leader may flip the first switch down, if the first switch is already down, then the leader will flip the second switch. The leader remembers how many times he has flipped the first switch down. Once the leader has flipped the first switch down 44 times, he announces that all have visited the room.

It does not matter how many times a prisoner has visited the room, in which order the prisoners were sent or even if the first switch was initially up. Once the leader has flipped the switch down 44 times then the leader knows everyone has visited the room. If the switch was initially down, then all 22 prisoners will flip the switch up twice. If the switch was initially up, then there will be one prisoner who only flips the switch up once and the rest will flip it up twice.

Soln to the Bridge and the Juggler Problem

London bridge is falling down; because a less intelligent juggler is trying his feat over the fragile bridge. This is about the juggler story (Where does the weight go?) which was posted earlier in this blog.

So, we had a classical riddle of a man crossing a bridge juggling all the way thereby reducing the effective weight acting on the bridge. We had to find a way to mathematically disprove this solution.

Note that when the man applies some force on a ball (to throw it up), the ball applies an equal and opposite force on him (downwards). This force gets transferred to the bridge as there is no acceleration of the man in the vertical direction. So, the total force acting on the bridge is the sum of the weight force of the man and the force applied by the ball.

( Let < signify "less than or equal to" and > signify "more than or equal to")

Let T be the time for which a ball remains in air. And let t be the time for which the ball remains in the hands of the man. Since the man has to handle 9 more balls before the first ball returns in his hands,
T > 9t

Let v be the velocity with which the ball leaves the man's hands. This is also the maximum velocity of the ball. Time required for the ball to reach the highest point is T/2. Hence,

v=gT/2

Let the man apply a constant force F on the ball in his hand for a period t while throwing it up. The ball comes down with a velocity v downwards and is thrown back up with a velocity v upwards. Hence the change in momentum of the ball is m(2v), where m is the mass of the ball. Hence,

F = 2.m.v/t + mg

=> 2mv = (F - mg).t

by the inequality of t,

(F - mg).T/9 > 2.m.v

Substituting the value of v,

(F - mg).T/9 > 2.m.(gT/2)

(F - mg)/9 > mg

F - mg > 9 mg

F > 10 mg

It implies that the force applied by each ball on the man would be greater than or equal to the weight of all ten balls. If one juggles, he is effectively putting more force on the bridge than he would otherwise have. Hence, juggling does no good. The strength of the bridge has to be at least 70.9 kg for that man to pass through with those balls; and even with that strength, it's not prudent to juggle.


:) crazy solution: The man might jog around for a couple of weeks and lose some weight :-)

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