Question

Sixteen players \[{s_1}\], \[{s_2}\],…, \[{s_{16}}\] ​ 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. The probability that "exactly one of the two players \[{s_1}\] and \[{s_2}\] ​ is among the eight winners" is
A.\[\dfrac{4}{{15}}\]
B.\[\dfrac{7}{{15}}\]
C.\[\dfrac{8}{{15}}\]
D.\[\dfrac{9}{{15}}\]

Answer 1

Hint: Here we have to find the probability that exactly one of the two players \[{s_1}\] and \[{s_2}\] ​wins the game. We will take two cases for finding the probabilities. In case 1, we will assume that the players \[{s_1}\] and \[{s_2}\] are in the same pair and in case 2, we will assume that the players \[{s_1}\] and \[{s_2}\]are in different pairs. Then we will first find the probability for these two cases separately. We will then add both the probabilities to get the required probability.

Complete step-by-step answer:
It is given that there are 16 players and they are divided into 8 pairs. We have to find the probability that exactly one of the two players \[{s_1}\] and \[{s_2}\] ​ wins the game.
Case 1: When \[{s_1}\] and \[{s_2}\] are in the same pair, then exactly one of them wins.
Therefore, total number of ways to divide 16 players into 8 pairs\[ = \dfrac{{16!}}{{{{\left( {2!} \right)}^8}8!}}\]
Since, \[{s_1}\] and \[{s_2}\] are in the same pair. So the number of ways to divide 14 players into 7 pairs
\[ = \dfrac{{14!}}{{{{\left( {2!} \right)}^7}7!}}\]
Probability of winning match by exactly one of the two players \[{s_1}\] and \[{s_2}\] \[ = \] number of ways to divide 14 players into 7 pairs \[ \div \] number of ways to divide 16 players into 8 pairs
Substituting all the values, we get
Probability of winning match by exactly one of the two players \[{s_1}\] and \[{s_2}\] \[ = \dfrac{{\dfrac{{14!}}{{{{\left( {2!} \right)}^7}7!}}}}{{\dfrac{{16!}}{{{{\left( {2!} \right)}^8}8!}}}}\]
Solving the factorials and simplifying the terms, we get
\[ \Rightarrow \] Probability of winning match by exactly one of the two players \[{s_1}\] and \[{s_2}\]\[ = \dfrac{1}{{15}}\]
Case 2: When \[{s_1}\] and \[{s_2}\] are in the different pair, then exactly one of them wins.
Therefore, total number of ways to divide 16 players into 8 pairs\[ = \dfrac{{16!}}{{{{\left( {2!} \right)}^8}8!}}\]
The number of ways to divide the players such that \[{s_1}\] and \[{s_2}\] are in the different pairs
\[ = \dfrac{{16!}}{{{{\left( {2!} \right)}^8}8!}} - \dfrac{{14!}}{{{{\left( {2!} \right)}^7}7!}}\]
Probability of winning match by exactly one of the two players \[{s_1}\] and \[{s_2}\] \[ = \dfrac{{\dfrac{{16!}}{{{{\left( {2!} \right)}^8}8!}} - \dfrac{{14!}}{{{{\left( {2!} \right)}^7}7!}}}}{{\dfrac{{16!}}{{{{\left( {2!} \right)}^8}8!}}}} \times \dfrac{1}{2} \times \dfrac{1}{2} \times 2\]
Solving the factorials and simplifying the terms, we get
\[ \Rightarrow \] Probability of winning match by exactly one of the two players \[{s_1}\] and \[{s_2}\] \[ = \dfrac{7}{{15}}\]
Therefore, total probability is equal to the sum of the probabilities obtained in case 1 and case 2.
Therefore,
Total probability \[ = \dfrac{1}{{15}} + \dfrac{7}{{15}} = \dfrac{8}{{15}}\]
Hence, the correct option is C.

Note: Factorial of any positive integer is defined as the multiplication of all the positive integers less than or equal to the given positive integers. Some important properties of factorial are as follows:-
(i)Factorial of zero is one.
(ii)Factorials are commonly used in permutations and combinations problems.
(iii)Factorials of negative integers are not defined.