On the school board a chairman is chosen. There are four candidates: \(A\), \(B\), \(C\) and \(D\). A special procedure is proposed – each member of the council writes down on a special sheet of candidates the order of his preferences. For example, the sequence \(ACDB\) means that the councilor puts \(A\) in the first place, does not object very much to \(C\), and believes that he is better than \(D\), but least of all would like to see \(B\). Being placed in first place gives the candidate 3 points, the second – 2 points, the third – 1 point, and the fourth - 0 points. After collecting all the sheets, the election commission summarizes the points for each candidate. The winner is the one who has the most points.
After the vote, \(C\) (who scored fewer points than everyone) withdrew his candidacy in connection with his transition to another school. They did not vote again, but simply crossed out \(B\) from all the leaflets. In each sheet there are three candidates left. Therefore, first place was worth 2 points, the second – 1 point, and the third – 0 points. The points were summed up anew.
Could it be that the candidate who previously had the most points, after the self-withdrawal of \(B\) received the fewest points?
In one box, there are two pies with mushrooms, in another box there are two with cherries and in the third one, there is one with mushrooms and one with cherries. The pies look and weigh the same, so it’s not known what is in each one. The grandson needs to take one pie to school. The grandmother wants to give him a pie with cherries, but she is confused herself and can only determine the filling by breaking the pie, but the grandson does not want a broken pie, he wants a whole one.
a) Show that the grandmother can act so that the probability of giving the grandson a whole pie with cherries will be equal to \(2/3\).
b) Is there a strategy in which the probability of giving the grandson a whole pie with cherries is higher than \(2/3\)?
There were 50 white and black crows sitting on a birch, and the number of black crows was not less than the number of whites. On the oak, there too were white and black crows, and there were 50 of them in total. On the oak, the number of black crows was also not less than the number of white ones. It could be that there was the same number of black and white crows, or maybe even there was one black crow less than white crows. One random crow flew from the birch to the oak, and after a while another random crow (maybe the same one) flew from the oak to the birch. Which is more probable: that the number of white crows on the birch is the same as it was at first, or that it has changed?
At the Antarctic station, there are \(n\) polar explorers, all of different ages. With the probability \(p\) between each two polar explorers, friendly relations are established, regardless of other sympathies or antipathies. When the winter season ends and it’s time to go home, in each pair of friends the senior gives the younger friend some advice. Find the mathematical expectation of the number of those who did not receive any advice.
In a tournament, 100 wrestlers are taking part, all of whom have different strengths. In any fight between two wrestlers, the one who is stronger always wins. In the first round the wrestlers broke into random pairs and fought each other. For the second round, the wrestlers once again broke into random pairs of rivals (it could be that some pairs will repeat). The prize is given to those who win both matches. Find:
a) the smallest possible number of tournament winners;
b) the mathematical expectation of the number of tournament winners.
At the ball, there were \(n\) married couples. In each pair, the husband and wife are of the same height, but there are no two pairs of the same height. The waltz begins, and all those who came to the ball randomly divide into pairs: each gentleman dances with a randomly chosen lady.
Find the mathematical expectation of the random variable \(X\), “the number of gentlemen who are shorter than their partners”.
On weekdays, the Scattered Scientist goes to work along the circle line on the London Underground from Cannon Street station to Edgware Road station, and in the evening he goes back (see the diagram).
Entering the station, the Scientist sits down on the first train that arrives. It is known that in both directions the trains run at approximately equal intervals, and along the northern route (via Farringdon) the train goes from Cannon Street to Edgware Road or back in 17 minutes, and along the southern route (via St James Park) – 11 minutes. According to an old habit, the scientist always calculates everything. Once he calculated that, from many years of observation:
– the train going counter-clockwise, comes to Edgware Road on average 1 minute 15 seconds after the train going clockwise arrives. The same is true for Cannon Street.
– on a trip from home to work the Scientist spends an average of 1 minute less time than a trip home from work.
Find the mathematical expectation of the interval between trains going in one direction.
A sequence consists of 19 ones and 49 zeros, arranged in a random order. We call the maximal subsequence of the same symbols a “group”. For example, in the sequence 110001001111 there are five groups: two ones, then three zeros, then one one, then two zeros and finally four ones. Find the mathematical expectation of the length of the first group.
There are \(n\) random vectors of the form \((y_1, y_2, y_3)\), where exactly one random coordinate is equal to 1, and the others are equal to 0. They are summed up. A random vector a with coordinates \((Y_1, Y_2, Y_3)\) is obtained.
a) Find the mathematical expectation of a random variable \(a^2\).
b) Prove that \(|a|\geq \frac{1}{3}\).
On one island, one tribe has a custom – during the ritual dance, the leader throws up three thin straight rods of the same length, connected in the likeness of the letter capital \(\pi\), \(\Pi\). The adjacent rods are connected by a short thread and therefore freely rotate relative to each other. The bars fall on the sand, forming a random figure. If it turns out that there is self-intersection (the first and third bars cross), then the tribe in the coming year are waiting for crop failures and all sorts of trouble. If there is no self-intersection, then the year will be successful – satisfactory and happy. Find the probability that in 2019, the rods will predict luck.