Question

Eight players \[{P_1},{P_2},.....{P_8}\] plays a knockout tournament it is known that whenever \[{P_i}\] and \[{P_j}\] play, the player \[{P_i}\] will win if \[i < j\] . Assume that the players are paired at random in each round. What is the probability that \[{P_4}\] reaches the final?
A. \[\dfrac{4}{{35}}\]
B. \[\dfrac{5}{{35}}\]
C. \[\dfrac{6}{{35}}\]
D. \[\dfrac{1}{{35}}\]

Answer 1

Hint: First of all find the number of ways in which \[{P_1},{P_2},.....{P_8}\] . can be paired in 4 pairs now do note that two players certainly reach the second round in between \[{P_1},{P_2},{P_3}\] . Use these things to solve the above question

Complete step-by-step answer:
The number of ways in which \[{P_1},{P_2},.....{P_8}\] can be paired in four pairs
\[\begin{array}{l}
\dfrac{1}{{4!}}({}^8{C_2})({}^6{C_2})({}^4{C_2})({}^2{C_2}) = \dfrac{1}{{4!}} \times \dfrac{{8!}}{{2!6!}} \times \dfrac{{6!}}{{2!4!}} \times \dfrac{{4!}}{{2!2!}} \times 1\\
 = \dfrac{1}{{4!}} \times \dfrac{{8 \times 7}}{{2! \times 1}} \times \dfrac{{6 \times 5}}{{2! \times 1}} \times \dfrac{{4 \times 3}}{{2! \times 1}}\\
 = \dfrac{{8 \times 7 \times 6 \times 5}}{{2 \times 2 \times 2 \times 2}}\\
 = 105
\end{array}\]
Now, at least two players certainly reach the second round in between \[{P_1},{P_2},{P_3}\] . And \[{P_4}\] can reach in final if exactly two players against each other in between \[{P_1},{P_2},{P_3}\] and remaining players will play against one of the players from \[{P_5},{P_6},{P_7},{P_8}\] and \[{P_4}\] play against one of the remaining three \[{P_5},{P_6},{P_7},{P_8}\]
This can be possible in \[{}^3{C_2} \times {}^4{C_1} \times {}^3{C_1} = 3 \times 4 \times 3 = 36ways\]
Therefore the probability that \[{P_4}\] and exactly one of the \[{P_5},{P_6},{P_7},{P_8}\] reach second round \[ = \dfrac{{36}}{{105}} = \dfrac{{12}}{{35}}\]
If \[{P_1},{P_i},{P_4},{P_j}\] where \[i = 2,3\& j = 5,6,7\] reach the second round then they can be paired in 2 pairs in \[\dfrac{1}{{2!}}\left( {{}^4{C_2}} \right)\left( {{}^2{C_2}} \right) = 3ways\]
But \[{P_4}\] will reach final round from the second \[ = \dfrac{1}{3}\]
Therefore the probability that \[{P_4}\] reach the final is \[\dfrac{{12}}{{35}} \times \dfrac{1}{3} = \dfrac{4}{{35}}\]

So, the correct answer is “Option A”.

Note: The calculation plays a very small part in this type of questions the key is to find the correct combinatorics logic in this question the rest is you can see as it is.