Eight players P:I, P2,. ..... P8 play a knock-out tournament. It is known that whenever the players Pi and Pj play; the player Pi will win if i < j. Assuming that the players are paired at random in each round, what is the probability that the player P4 reaches the final ?
Ms.Sue?? Steve sir?? Teachers??
How can player Pi win, if "i<j"?
See, it goes like this. Say for example P3 plays against P5 (i=3, j=5). So according to the question P3 will win because 3<5.
I am extremely sorry. The question should have been as follows.
*Eight players P1,P2..,P8 play 3 sets of a knockout tournament....*
To find the probability that player P4 reaches the final in the knock-out tournament, we need to consider the possible outcomes in each round.
Let's analyze the rounds step by step:
1. Round 1:
- Player P4 needs to win their match to advance to the next round.
The probability of P4 winning in the first round is 1/2 since there are two equally likely outcomes (win or lose).
2. Round 2:
- P4 needs to win in the second round to advance to the next stage.
- In this round, P4 will be paired against one of the remaining 7 players: P1, P2, P3, P5, P6, P7, or P8.
The probability of winning the second round is 1/7 since there are seven equally likely outcomes (each potential opponent).
3. Round 3:
- P4 needs to win in the third round to reach the final.
- In this round, P4 will be paired against one of the remaining 3 players, determined by the outcome of the previous rounds.
The probability of winning the third round is 1/3 since there are three remaining players, and each outcome is equally likely.
4. Final:
- P4 has reached the final, so it does not matter who they play against.
The probability of reaching the final is 1.
To find the overall probability, we multiply the probabilities of each round:
(1/2) * (1/7) * (1/3) * 1 = 1/42
Therefore, the probability that player P4 reaches the final is 1/42.