Collection of Maths Problems

Toy train

Task number: 3539

• Variant

  What is the probability that the train will hold together in a given order (you may not turn the wagons around or rearrange them) if the magnets' polarities are assigned at random in the factory?

• Solution

  At each junction the probability is \(\frac12\) that the magnets have compatible polarities.

• Answer

  The probability is \(\frac18\).

• Variant

  What is the probability that a train can be constructed in at least one possible order (you may turn the wagons around or rearrange them) if the magnets' polarities are assigned at random in the factory?

• Hint

  Try to first find some condition that must be fulfilled for the train to be assembled.

• Solution

  The set of carriages has seven positions for magnets, so there are \(2^7\) possible polarity assignments.

  If it is possible to connect the train, 4 ends will have positive polarity and 3 will have negative polarity, or conversely. This is however a sufficient condition: a wagon whose ends have different polarities can be placed anywhere, and a wagon whose ends have the same polarity, e.g. negative, can be paired with a wagon with positive polarity and this pair can be placed anywhere, or possibly paired with the locomotive.

  The total such possible arrangements of the magnets is \(2\binom{7}{3}\).

• Answer

  The probability that a random assignment of magnets can be connected is \(\frac{35}{64}\).

• Variant

  What is the probability that a train can be constructed from the wagons in every possible order (you may turn the wagons around in any given order) if the magnets' polarities are assigned at random in the factory?

• Solution

  Here two cases arise:

  • All wagons have ends with opposite polarities. Then there are two choices for each wagon and also two for the locomotive – \(2^4\) possibilities.
  • Exactly one wagon has ends with the same polarity and the locomotive has the opposite polarity, so there are two choices for the given wagon and also two for each of the other wagons. The polarity of the locomotive is given uniquely – \(3{\cdot} 2^3\) possibilities.

  Observe that if the set contained two wagons each of whose ends had the same polarity, the either it would not be possible to connect them to each other, or it would not be possible to connect one of them to the locomotive.

• Answer

  The probability that it is possible to connect every sequence of wagons is \(\frac{5}{16}\).

• Variant

  Generalize the preceding variants for \(n\) wagons.

• Answer

  In the first case the general expression is \(\frac{1}{2^n}\), in the second \(\frac{\binom{2n+1}{n}}{2^{2n}}\) and in the third \(\frac{3}{2^{n+1}}\).