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

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

\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.