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Eight players P1, P2, P3, .............P8 play a knockout 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 players P4 reaches the final
  • a)
    4/35
  • b)
    5/35
  • c)
    6/35
  • d)
    9/35
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Eight players P1, P2, P3, .............P8 play a knockout tournament. ...
Given that if Pi, Pj play with i < j, then Pi will win. For the first round, P4 should be paired with any one from P5 to P8. It can be done in 4C1  ways. Then P4 to be the finalist, at least one player from P5 to P8 should reach in the second round. Therefor, one pair should be from remaining 3 from P\ to P8 in 3C2 Then favourable pairing in first round is 4C1 3C2 3C2. Then in the 2nd round, we have four players.
Favourable ways in 1. Now total possible pairings is
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Most Upvoted Answer
Eight players P1, P2, P3, .............P8 play a knockout tournament. ...
To find the probability that player P4 reaches the final, we need to determine the number of possible paths that lead to P4 reaching the final and divide it by the total number of possible paths.

Let's consider the possible paths for P4 to reach the final:

1. P4 wins all matches:
- In the first round, P4 can be paired with any of the remaining 7 players.
- In the second round, P4 can be paired with any of the remaining 3 players.
- In the semi-final, P4 can be paired with any of the remaining 2 players.
- In the final, P4 will reach the final as there is only 1 player left.
- Therefore, the number of paths where P4 wins all matches is 7 * 3 * 2 * 1 = 42.

2. P4 loses in the first round and then wins all matches:
- In the first round, P4 can be paired with any of the remaining 7 players.
- In the second round, P4 can be paired with any of the remaining 3 players.
- In the semi-final, P4 can be paired with any of the remaining 2 players.
- In the final, P4 will reach the final as there is only 1 player left.
- Therefore, the number of paths where P4 loses in the first round and then wins all matches is 7 * 3 * 2 * 1 = 42.

3. P4 loses in the first two rounds and then wins all matches:
- In the first round, P4 can be paired with any of the remaining 7 players.
- In the second round, P4 can be paired with any of the remaining 3 players.
- In the semi-final, P4 can be paired with any of the remaining 2 players.
- In the final, P4 will reach the final as there is only 1 player left.
- Therefore, the number of paths where P4 loses in the first two rounds and then wins all matches is 7 * 3 * 2 * 1 = 42.

4. P4 loses in the first three rounds and then wins all matches:
- In the first round, P4 can be paired with any of the remaining 7 players.
- In the second round, P4 can be paired with any of the remaining 3 players.
- In the semi-final, P4 can be paired with any of the remaining 2 players.
- In the final, P4 will reach the final as there is only 1 player left.
- Therefore, the number of paths where P4 loses in the first three rounds and then wins all matches is 7 * 3 * 2 * 1 = 42.

Summing up all the paths, we have a total of 42 + 42 + 42 + 42 = 168 paths where P4 reaches the final.

The total number of possible paths in the tournament can be calculated as follows:

- In the first round, there are 8 players, so the number of possible paths is 8! (factorial of 8).
- In the second round, there are 4 players left, so the number of possible paths is 4!.
- In the semi-final, there are 2
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Eight players P1, P2, P3, .............P8 play a knockout 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 players P4 reaches the finala)4/35b)5/35c)6/35d)9/35Correct answer is option 'A'. Can you explain this answer?
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