Waste less time on Facebook — follow Brilliant.
×

Probability Question!! Please Help..!!!

Eight Player \( P_{1},P_{2},P_{3},P_{4}.......P_{8} \) are going to play a Knock-Out Tournament.It is known that whenever \( P_{i} \) and \( P_{j} \) plays , The Player \( P_{i} \) wins if i<j . Assuming that Players are paired at random in each round, What is the probability that player \( P_{4} \) reaches the final ??

Also Find the probability of all the other Players in reaching the final ?

What should be sum of all the individual probabilities of each player in reaching the final ?

Please help me solve my doubt..!!!

Note by Rushi Rokad
4 years, 1 month ago

No vote yet
8 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

For first question, I think \(P\) \(=\) \(\frac{4}{35}\).

Reason: Let, integers from \(1\) to \(8\) represent respective players. Firstly, in order for \(P_{4}\) to win, the player \(4\) has to be paired with a higher number with is one of these: \(5\), \(6\), \(7\) or \(8\). This can be done with a probability \(\frac{1}{4}\).

Secondly, in order to win, two large numbers greater than \(4\) has to be paired to ensure there is a larger number to win from when 4 reaches in the round before semi-final. This can be done with a probability \(\frac{^3C_2}{^6C_2}\).

Finally, making use of Bayes Theorem, \(P\) \(=\) \(\frac{1}{4}\times\frac{^3C_2}{^6C_2}\) = \(\frac{4}{35}\).

Though I am a bit hazy about the answer due to the random effect (players are shuffled after every round).

Lokesh Sharma - 4 years, 1 month ago

Log in to reply

No..I Think you are correct.

Rushi Rokad - 4 years, 1 month ago

Log in to reply

I was thinking about this problem and i think I've figured it out! P1 has a 7/7 chance, or %100 P2 has a 6/7 chance, or about %86 P3 has a 5/7 chance, or about %71 P4 has a 4/7 chance, or about %51 P5 has a 3/7 chance, or about %43 P6 has a 2/7 chance, or about %29 P7 has a 1/7 chance, or about %14 P8 has a 0/7 chance, or %0 these are all decimals except for P1 and P8

Billy Coleman - 4 years, 1 month ago

Log in to reply

The chance of \( P_{7} \) and \( P_{6} \) in reaching the final is also zero...The probabilities you have given are for Players to reach the semifinal and not the finals..

Rushi Rokad - 4 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...