In a knock out chess tournament, eight players $${P_1},\,{P_2},......,\,{P_8}$$ participated. It is known that whenever the players $${P_i}$$ and $${P_j}$$ play, the players $${P_i}$$ will win $$j$$ 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.
$$\frac{{31}}{{35}}$$
B.
$$\frac{4}{{35}}$$
C.
$$\frac{8}{{35}}$$
D.
none of these
Answer :
$$\frac{4}{{35}}$$
Solution :
Let us divide the players into two pools $$A$$ and $$B$$ each containing $$4$$ players.
Let $${P_4}$$ be in pool $$A$$. Now $${P_4}$$ will reach the final if we fill the remaining three of pool $$A$$ by any of $${P_5},\,{P_6},\,{P_7}{\text{ or }}{P_8}$$
$$\therefore $$ Probability is $$\frac{{{}^4{C_3}}}{{{}^7{C_3}}} = \frac{{4.3.2}}{{7.6.5}} = \frac{4}{{35}}$$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$