SATHEE: Probability 4 Question 21

Probability 4 Question 21

21. Eight players $P_{1}, P_{2}, \dots, P_{8}$ 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) $\frac{8}{15}$ 24. $\frac{24}{29}$

Solution:

The number of ways in which $P_{1}, P_{2}, \dots, P_{8}$ can be paired in four pairs

$\begin{aligned} = \frac{1}{4!} [(^{8} C_{2}) (^{6} C_{2}) (^{4} C_{2}) (^{2} C_{2})] \\ = \frac{1}{4!} \times \frac{8!}{2! 6!} \times \frac{6!}{2! 4!} \times \frac{4!}{2! 2!} \times 1 \\ = \frac{1}{4!} \times \frac{8 \times 7}{2! \times 1} \times \frac{6 \times 5}{2! \times 1} \times \frac{4 \times 3}{2! \times 1} = \frac{8 \times 7 \times 6 \times 5}{2 \cdot 2 \cdot 2 \cdot 2} = 105 \end{aligned}$

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 $P_{5} \dots P_{8}$ .

This can be possible in

$^{3} C_{2} \times^{4} C_{1} \times^{3} C_{1} = 3 \cdot 4 \cdot 3 = 36 ways$

$∴$ Probability that $P_{4}$ and exactly one of $P_{5} \dots P_{8}$ 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 $\frac{1}{2!} (^{4} C_{2}) (^{2} C_{2}) = 3$ ways. But $P_{4}$ will reach the final, if $P_{1}$ plays against $P_{i}$ and $P_{4}$ plays against $P_{j}$ .