Question

A chess game between \[X\] and \[Y\] is won by whoever first wins a total of 2 games. \[X\]’s chances of winning, drawing or losing a particular game are\[\dfrac{1}{6}, \dfrac{1}{3}, \dfrac{1}{2}\] . The games are independent. Then find the probability that \[Y\] wins that match in the \[{4^{th}}\] game.
A. \[\dfrac{1}{6}\]
B. \[\dfrac{1}{4}\]
C. \[\dfrac{1}{2}\]
D. None of these

Answer 1

Hint: Use the given probabilities of winning, drawing, or losing a particular game of the player \[X\] to calculate the probabilities of winning, drawing, or losing a particular game of the player \[Y\]. Then calculate the possible ways where player \[Y\] wins the match in the \[{4^{th}}\] game. In the end, substitute the values of the probabilities in that equation and simplify it to get the required answer.

Formula used:
Combination formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Complete step by step solution:
The given probabilities of winning, drawing, or losing a particular game of player \[X\] are \[\dfrac{1}{6}, \dfrac{1}{3}, \dfrac{1}{2}\] respectively.

Let’s calculate the probabilities of winning, drawing, or losing a particular game of the player \[Y\].
When the player \[X\] wins, the player \[Y\] loses the match.
So, \[P\left( {Y loose} \right) = \dfrac{1}{6}\]
When the player \[X\] loses, the player \[Y\] wins the match.
So, \[P\left( {Y win} \right) = \dfrac{1}{2}\]
And the probability of drawing the match is, \[P\left( {Draw} \right) = \dfrac{1}{3}\].
Let \[P\left( E \right)\] be the probability that \[Y\] wins that match in the \[{4^{th}}\] game.
So, the player \[Y\] wins that match in the \[{4^{th}}\] game is fixed.
The possible ways are,
\[\left( {X win} \right) \left( {Draw} \right) \left( {Y win} \right)\left( {Y win} \right) \] and \[\left( {Draw} \right) \left( {Draw} \right) \left( {Y win} \right)\left( {Y win} \right)\]
In a first way, the results of the first 3 games can be arranged in \[3!\] ways.
In a second way, the possibility of the drawing the game occurs 2 times.
Therefore, the probability that \[Y\] wins that match in the \[{4^{th}}\] game is
\[P\left( E \right) = 3! \times P\left( {\left( {X win} \right) \left( {Draw} \right) \left( {Y win} \right)\left( {Y win} \right) } \right) + {}^3{C_2} \times P\left( {\left( {Draw} \right) \left( {Draw} \right) \left( {Y win} \right)\left( {Y win} \right) } \right)\]
\[ \Rightarrow P\left( E \right) = 6 \times \left( {\dfrac{1}{6} \times \dfrac{1}{3} \times \dfrac{1}{2} \times \dfrac{1}{2}} \right) + \dfrac{{3!}}{{1! \times 2!}} \times \left( {\dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{2} \times \dfrac{1}{2} } \right)\]
\[ \Rightarrow P\left( E \right) = \dfrac{1}{{12}} + \dfrac{1}{{12}}\]
\[ \Rightarrow P\left( E \right) = \dfrac{2}{{12}}\]
\[ \Rightarrow P\left( E \right) = \dfrac{1}{6}\]
Hence the correct option is A.

Note: Always remember to calculate the number of ways of arranging the possible result of each game. Students often forget to calculate the arrangements of the results and directly calculate the probability.