Probability Ques 109
Sixteen players $S _1, S _2, \ldots, S _{16}$ 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: 109.(i)$\frac{1}{2}$ (ii)$\frac{8}{15}$
Solution:
Formula:
- (i) Probability of $S _1$ to be among the eight winners $=\left(\right.$ Probability of $S _1$ being a pair $)$ $\times$ (Probability of $S _1$ winning in the group)
$=1 \times \frac{1}{2}=\frac{1}{2} \quad$ [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
$=\left(\right.$ 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.
$\therefore \quad n(E)=\frac{(14) !}{(2 !)^{7} \cdot 7 !}$ and $n(S)=\frac{(16) !}{(2 !)^{8} \cdot 8 !}$
$\therefore$ 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) $=1$
$\therefore$ 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} $
Required Probability $=\frac{1}{15} + \frac{7}{15} = \frac{8}{15}$
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