Depending on the meaning of the question, the value of x can be 3, 4, 5.
When X=3, the probability is1/c (3,5) =110.
When X=4, the probability is c (2,3)/c (3,5) = 3/1 0 (the maximum value is 4, and the other two are randomly selected from1,2,3).
When X=5, the probability is c (2,4)/c (3,5) = 6/1 0 (the maximum value is 5, and the other two are randomly selected from1,2,3,4).
So look forward to it
e(X)= 1/ 10 X 3+3/ 10 X 4+6/ 10 X 5
=45/ 10
=4.5
Extended data:
Historical story of mathematical expectation
/kloc-in the 0 th and 7 th centuries, a gambler challenged Pascal, a famous French mathematician, and gave him a topic: A and B gamble, and the chances of winning are equal. The rules of the game are that the winner is the one who wins three games first, and the winner who wins five games can get a reward of 100 francs. In the fourth game, A won two games and B won one. At this time, the game is suspended for some reason, so how to distribute the 65,438+000 francs is fair.
With the knowledge of probability theory, it is not difficult to know that A is likely to win and B is unlikely to win.
Because the probability of A losing the last two games is only (1/2) × (1/2) =1/4, that is to say, the probability of A winning the last two games or any of them is1-(1/4) = 3. However, if B expects to win 100 francs, it must beat A in the last two games. The probability of B winning the last two games in a row is (1/2) * (1/2) =1/4, that is, B has a 25% probability of winning 100 francs.
It can be seen that although the game can't be played again, according to the above possibilities, the objective expectations of both parties for the final victory are 75% and 25% respectively, so Party A should get 100 * 75% = 75 (francs) and Party B should get 100×25% = 25 (francs). The word "expectation" appeared in this story, from which mathematical expectation came.