00:01
A tennis tournament is arranged for 2 raised to n number of players.
00:06
It is organized as a knocked out and so that only the winners in any given round proceed to the next round.
00:17
Opponents in each round except the final are drawn at random and in any match either player has a probability 1 -2 of winning.
00:26
1 by 2 of winning it has a probability.
00:29
2 players are chosen at a random, let p1 and p2 be any 2 players and they are chosen at random before the start of a first round.
00:39
Find the probability that each player play in the first round.
00:45
So we will find this for the first round of a play.
00:51
So calculating this, let the 2 players be p1 and p2.
00:57
Then in the first part of a question, once p1 has been given a slot, then 2 raised to n -1 slots are remaining for the player p2 in only one of which will he or she play p1.
01:15
P1 has only one slot to play and 2 raised to n -1 slots are available for p2.
01:22
Now the probability of player p1 playing p2 is therefore given by 1 upon 2 raised to n -1.
01:33
Since the probability of p1 is 1 and probability of p2 is 2 raised to n -1.
01:39
This is the required answer for first part of a question.
01:42
Here to note that this works only for n equal to 1 and for n equal to 2.
01:48
Now in the second part of a question, we have to calculate the probability that the 2 players play in the final round.
01:59
For this, we will take the long way and the short way.
02:03
Two cases here will be there.
02:05
So for the long way to meet in the final round, p1 and p2 must each win each round before the final and must also not meet before the final.
02:16
This must be the case.
02:17
The probability that p1 will not meet the p2 in the first round and that they both win their first round matches is respectively 1 -1 upon 2 raised to power n -1 multiplied by 1 upon 2 square.
02:35
That is, it must be 1 upon 2 multiplied by 2 raised to n -1 minus 1 divided by 2 raised to n -1.
02:47
The probability that they win each round and do not meet before the final for n -1 rounds, for n -1 rounds, they win for n before, this is 1 by 2 multiplied by 2 raised to power n -1 minus 1 divided by 2 raised to power n -1 multiplied by 1 upon 2 into 2 raised to n -2 minus 1 divided by 2 raised to n -1 minus 1 and so on.
03:24
This process will continue.
03:26
It is for only 1 for the first round and for n -1 round, this first, second and up to n -1 rounds will multiply here...