Question

Eight players $P_1,P_2,\cdots ,P_8$ paly a knock - out tournament. It is known that whenever the players $P_{i}and{P}_j$ play, the player $P_i$ will win if i < j. Assuming that the players are paired at random in each round, what is the probability that the player $P_4$ reaches the final?

• A. $\frac{3}{35}$
• B. $\frac{4}{35}$
• C. $\frac{1}{5}$
• D. $\frac{12}{35}$

Answer 1

The correct option is B $\frac{4}{35}$

The number of ways in which $P_1,P_2,\cdots ,P_8$ can be paired in four pairs
$=\frac{1}{4!}\left[{(}^8{C}_2\right){(}^6{C}_2\left){(}^4{C}_2\right){(}^2{C}_2\left)\right]=\frac{1}{4!}\times \frac{8!}{2!6!}\times \frac{6!}{2!4!}\times \frac{4!}{2!2!}\times 1=\frac{1}{4!}\times \frac{8\times 7}{2!\times 1}\times \frac{6\times 5}{2!\times 1}\times \frac{4\times 3}{2!\times 1}=\frac{8\times 7\times 6\times 5}{\mathrm{2.2.2.2}}=105$
Now, atleast two players certainly reach the second round between $P_1,P_2,P_{3}and{P}_4.P_4$ can reach in final if exactly two players play against each other between $P_1,P_2,P_3$ and remaining player will play against one of the players from $P_5,P_6,P_7,P_{8}and{P}_4$ plays against one of the remaining three from $P_5\cdots P_8$.
This can be possible in
${}^3{C}_2{\times }^4{C}_1{\times }^3{C}_1=\mathrm{3.4.3}=36ways$
$\therefore$ Probability that $P_4$ and exactly one of $P_5\cdots P_8$ reach second round
$=\frac{36}{105}=\frac{12}{35}$
If $P_1,P_i,P_{4}and{P}_j$, where i = 2 or 3 and j = 5 or 6 or 7 reach the second round, then they can be paired in 2 pairs in $\frac{1}{2!}{(}^4{C}_2\left){(}^2{C}_2\right)=3$ ways. But $P_4$ will reach the final, if $P_1$ plays against $P_{i}and{P}_4$ plays against $P_j$
Hence, the probability that $P_4$ will reach the final round from the second $=\frac{1}{3}$
$\therefore$ Probability that $P_4$ will reach the final is $\frac{12}{35}\times \frac{1}{3}=\frac{4}{35}$