SATHEE: Probability 4 Question 23

Probability 4 Question 23

23. Sixteen players $S_{1}, S_{2}, \dots, S_{16}$ play in a tournament. They are divided into eight pairs at random from each pair a winner is decided on the basis of a game played between the two players of the pair. Assume that all the players are of equal strength.

(i) Find the probability that the player $S_{1}$ is among the eight winners.

(ii) Find the probability that exactly one of the two players $S_{1}$ and $S_{2}$ is among the eight winners.

(1997C, 5M)

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Answer:

Correct Answer: 23. $\frac{99}{1900}$

Solution:

(i) Probability of $S_{1}$ to be among the eight winners $= ($ Probability of $S_{1}$ being a pair $)$ $\times$ (Probability of $S_{1}$ winning in the group) $= 1 \times \frac{1}{2} = \frac{1}{2}$ [since, $S_{1}$ is definitely in a group]

(ii) If $S_{1}$ and $S_{2}$ are in the same pair, then exactly one wins.

If $S_{1}$ and $S_{2}$ are in two pairs separately, then exactly one of $S_{1}$ and $S_{2}$ will be among the eight winners. If $S_{1}$ wins and $S_{2}$ loses or $S_{1}$ loses and $S_{2}$ wins.

Now, the probability of $S_{1}, S_{2}$ being in the same pair and one wins

$= ($ Probability of $S_{1}, S_{2}$ being the same pair $)$

$\times$ (Probability of anyone winning in the pair). and the probability of $S_{1}, S_{2}$ being the same pair

$= \frac{n (E)}{n (S)}$

where, $n (E) =$ the number of ways in which 16 persons can be divided in 8 pairs.

$∴ n (E) = \frac{(14)!}{(2!)^{7} \cdot 7!}$ and $n (S) = \frac{(16)!}{(2!)^{8} \cdot 8!}$

$∴$ Probability of $S_{1}$ and $S_{2}$ being in the same pair

$= \frac{(14)! \cdot (2!)^{8} \cdot 8!}{(2!)^{7} \cdot 7! \cdot (16)!} = \frac{1}{15}$

The probability of any one wining in the pairs of $S_{1}, S_{2} = P$ (certain event) $= 1$

$∴$ The pairs of $S_{1}, S_{2}$ being in two pairs separately and $S_{1}$ wins, $S_{2}$ loses +The probability of $S_{1}, S_{2}$ being in two pairs separately and $S_{1}$ loses, $S_{2}$ wins.

$\begin{aligned} = 1 - \frac{\frac{(14)!}{(2!)^{7} \cdot 7!}}{\frac{(16)!}{(2!)^{8} \cdot 8!}} \times \frac{1}{2} \times \frac{1}{2} + 1 - \frac{\frac{(14)!}{(2!)^{7} \cdot 7!}}{\frac{(16)!}{(2!)^{8} \cdot 8!}} \times \frac{1}{2} \times \frac{1}{2} \\ = \frac{1}{2} \times \frac{14 \times (14)!}{15 \times (14)!} = \frac{7}{15} \end{aligned}$