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In a boxing tournament 2n equally skilled players P1,P2,P3...........P$2^{n}$, are participating.

In each round players are divided in pairs at random and winner from each pair moves in next round. If P5 reaches the semifinals then what is the probability that P1 wins the tournament?

A. $(\frac{1}{2})^{logn}$

B. $\frac{2^{logn-1}}{2^{n}-1}$

C. $\frac{3}{4} * \frac{1}{2^{n}-1}$

D. $\frac{7}{8} * \frac{1}{2^{n}-1}$

asked in Probability by Veteran (10.3k points)
edited by | 208 views

1 Answer

+7 votes
Best answer

Lets take an example.

Suppose there 8 participants P1, P2..................P7, P8.

 

Now, they are saying find the probablity that P wins the tournamant given that P5 reaches semi-finals. 

So, sample space is P5, P, Pj , Pk (i.e. P5, P, Pj , Pare semi-finalists).

 

Now, we would like to find the probablity that P1 wins the tournament given the above sample space.

This means that P1 should reach the semifinals (i.e. P1 should be either one of the P, Pj , Pk).

 

Now, P(P1 is either one of the P, Pj , Pk)

= ( select any two for Pj , Pk  ) /  ( select any three for P, Pj , Pk)

= $\frac{\binom{6}{2}}{\binom{7}{3}}$          .......................................(1)

= $\frac{3}{7}$          

 

Now, we have got P1 in semifinals. So, we need to find probablity that P1 wins the tournament given that semifinalists are P5, P1 , Pj , Pk

P(P1 wins the tournament given that P5, P1 , Pj , Pare semi-finalists) =  $\frac{1}{4}$   .................(2)

 

$\therefore$ P(P1 wins the tournament / P5 reaches semis) = $\frac{1}{4}$ *  $\frac{3}{7}$ =  $\frac{3}{4}$ *  $\frac{1}{7}$

 

This answer matches with answer B.

----------------------------------------------------------------------

Lets generalize the answer.

(A) = $\frac{\binom{2^{n}-2}{2}}{\binom{2^{n}-1}{3}}$       ...........from statement(1) above

(B) = $\frac{1}{4}$   .......................from statement(2) above

 

$\therefore$ P(P1 wins the tournament / P5 reaches semis) = (B) * (A) = $\frac{3}{4} * \frac{1}{2^{n}-1}$

 

answered by Veteran (18k points)
selected by

Now, P(P1 is either one of the P, Pj , Pk)

= ( select any two for Pj , Pk  ) /  ( select any three for P, Pj , Pk)

Please explain this part once more 

The initial sample space is P5, Pi, Pj, Pk because we need to find probablity that P1 wins tournament given P5 reaches semis, right? I hope this clear.

Now, when will P1 win tournament? This is possible if P1 also reaches semis, right?

So, if we want to find the P(P1 wins tournament/ P5 reaches semis), then P1 has to find place in players P5, Pi, Pj, Pk , right?

So, P1 can be Pi or Pj or Pk.

------------------------------------------------------------------------------------

So, whats the probablity that P1 is Pi or Pj or Pk?  

the sample space is (2n-1)C3  since excluding P1, you can select any 3 players from remaining ones in place of Pi, Pj and Pk.

Favourable events are those where P1 is also in semis alongwith P5 i.e. P1, P5, Pj, Pk  for some j,k.

So, total favourable events = (2n-2)C2  i.e you are choosing for Pj and Pk.

Thats it.

 

Crystal Clear now!! Thanks!

@GateSet  which test series ? 

Testbook
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OK then keep posting good QS
Surely!!

@Debashish.

OK then keep posting good QS

Good comment ;)

But, really @Gateset that was wonerful question.

 

@sushant,though u solved perfectly,still not getting this part:-(

Now, P(P1 is either one of the P, Pj , Pk)

= ( select any two for Pj , Pk  ) /  ( select any three for P, Pj , Pk)

 

quite good explanation. @sushant! keep it up! and akriti just take a small sample space of maybe 5 players and follow his comment next to his answer. you will get it!
@Akriti. I hope you got the initial sample space P5, Pi, Pj, Pk............(1)

Now, In order that P1 wins tournament, P1 must also reach semis.

So, this is situation: P1, P5, Pj, Pk..................(2)

 

So, whats the probablity of situation in statement (2) given situation in statement (2) is sample space?

So, sample space = 7C3 since you need to select Pi,Pj,Pk

 

Events that account for situation in statement (2) = 6C2

since, you need to select Pj,Pk

 

So, prob = 6C2 / 7C3

@sushant,you are finding the probability that P1 is in semis-right?

then you need to select 3 personns for semis which can be done in 7C3 ways..i got till here.

but how favourable cases are 6C2??does this mean,you have chosen P1 and you are selectig rest 2 in semis..am i right??

Yup. As you know, whenever you are consider favourable events, you should assume that situation.


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