Probability 4 Question 21
21. Eight players $P _1, P _2, \ldots, P _8$ play 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?
$(1999,10 M)$
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Answer:
Correct Answer: 21. (i) $\frac{1}{2}$ (ii) $\frac{8}{15} \quad$ 24. $\frac{24}{29}$
Solution:
- The number of ways in which $P _1, P _2, \ldots, P _8$ can be paired in four pairs
$$ \begin{aligned} & =\frac{1}{4 !}\left[\left({ }^{8} C _2\right)\left({ }^{6} C _2\right)\left({ }^{4} C _2\right)\left({ }^{2} C _2\right)\right] \\ & =\frac{1}{4 !} \times \frac{8 !}{2 ! 6 !} \times \frac{6 !}{2 ! 4 !} \times \frac{4 !}{2 ! 2 !} \times 1 \\ & =\frac{1}{4 !} \times \frac{8 \times 7}{2 ! \times 1} \times \frac{6 \times 5}{2 ! \times 1} \times \frac{4 \times 3}{2 ! \times 1}=\frac{8 \times 7 \times 6 \times 5}{2 \cdot 2 \cdot 2 \cdot 2}=105 \end{aligned} $$
Now, atleast two players certainly reach the second round between $P _1, P _2$ and $P _3$ and $P _4$ 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 $P _5 \ldots P _8$.
This can be possible in
$$ { }^{3} C _2 \times{ }^{4} C _1 \times{ }^{3} C _1=3 \cdot 4 \cdot 3=36 \text { ways } $$
$\therefore$ Probability that $P _4$ and exactly one of $P _5 \ldots P _8$ reach second round $\quad=\frac{36}{105}=\frac{12}{35}$
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 $\frac{1}{2 !}\left({ }^{4} C _2\right)\left({ }^{2} C _2\right)=3$ 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 $\frac{12}{35} \times \frac{1}{3}=\frac{4}{35}$.
