Question

A and B play a match to be decided as soon as either has won two games. The chance of either winning a game is \[\dfrac{1}{{20}}\] and of its being drawn is \[\dfrac{9}{{10}}\]. What is the chance that the match is finished in 10 or less games?
A.\[0.18\] approx.
B.\[0.16\]approx.
C.\[0.17\]approx.
D.\[0.15\] approx.

Answer 1

Hint: Here, we will find the probability of the match not being completed in 10 or less games by taking the different cases possible that could be the reason for extending the matches. We will then subtract the obtained probability from the total probability i.e. 1 and find the required probability.
Formula Used:
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], where \[n\] is the total number of elements and \[r\] is the number of elements to be selected.

Complete step-by-step answer:
According to the question,
The probability of either winning a game is \[\dfrac{1}{{20}}\].
The probability of the game being drawn is \[\dfrac{9}{{10}}\].
We are required to find the probability that the match is finished in 10 or less games.
Let us assume that the match is not finished in 10 games.
Now, this is possible only when the following cases happen:
All games are drawn: If all the games are drawn then, neither A nor B will be able to win 2 games hence, the match will be stretched for more than 10 games.
Now, it is given in the question that the probability of one game being drawn is \[\dfrac{9}{{10}}\]
Hence,
The probability of 10 games being drawn \[ = {\left( {\dfrac{9}{{10}}} \right)^{10}}\]…………………………\[\left( 1 \right)\]
A and B each wins 1 game and the other 8 games are drawn: In this case also, since neither A nor B will be able to win 2 games hence, the match will be stretched for more than 10 games.
Hence, the probability of winning of A and B, 1 game each \[ = {\left( {\dfrac{1}{{20}}} \right)^2}\]
And,
The probability of the other 8 games being drawn \[ = {\left( {\dfrac{9}{{10}}} \right)^8}\]
Also, in this case, we have to use combinations to select that out of the 10 games, which 2 games are the one in which either A or B had won.
Therefore, using the formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], we get
\[{}^{10}{C_2} = \dfrac{{10!}}{{2!\left( {10 - 2} \right)!}}\]
Subtracting the terms in the denominator, we get
\[ \Rightarrow {}^{10}{C_2} = \dfrac{{10!}}{{2!8!}}\]
Computing the factorial, we get
\[ \Rightarrow {}^{10}{C_2} = \dfrac{{10 \times 9}}{2} = 45\]
Also, they can be arranged in \[2! = 2\]ways

Hence, the total probability that A and B each wins 1 game and the other 8 games are drawn will be \[ = 2 \times {}^{10}{C_2} \times {\left( {\dfrac{9}{{10}}} \right)^8} \times {\left( {\dfrac{1}{{20}}} \right)^2}\]…………………………….\[\left( 2 \right)\]
A or B wins one game and the other 9 games are drawn.
In this case, we have to use combinations to select that out of the 10 games, which game is the one in which either A or B has won.
Therefore, using the formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], we get,

\[{}^{10}{C_2} = \dfrac{{10!}}{{1!\left( {10 - 1} \right)!}}\]
Subtracting the terms in the denominator, we get
\[ \Rightarrow {}^{10}{C_2} = \dfrac{{10!}}{{1!9!}}\]
Computing the factorial, we get
\[ \Rightarrow {}^{10}{C_2} = 10\]

Also, there will be 2 cases, one in which A had won and the other in which B had won.
Now, the probability of either of them winning 1 game is \[\dfrac{1}{{20}}\]
And, the probability of the other 9 games being drawn \[ = {\left( {\dfrac{9}{{10}}} \right)^9}\]
Hence, the total probability that either A or B wins 1 game and the other 9 games are drawn will be \[ = 2 \times {}^{10}{C_1} \times {\left( {\dfrac{9}{{10}}} \right)^9} \times \left( {\dfrac{1}{{20}}} \right)\]…………………………….\[\left( 3 \right)\]
Now, since these three cases are mutually exclusive.
Hence, the total probability of the match not being finished in 10 games will be found out by adding all the 3 cases.
 Adding equation \[\left( 1 \right)\], \[\left( 2 \right)\] and \[\left( 3 \right)\], we get
Required Probability \[ = {\left( {\dfrac{9}{{10}}} \right)^{10}} + 2 \times {}^{10}{C_2} \times {\left( {\dfrac{9}{{10}}} \right)^8} \times {\left( {\dfrac{1}{{20}}} \right)^2} + 2 \times {}^{10}{C_1} \times {\left( {\dfrac{9}{{10}}} \right)^9} \times \left( {\dfrac{1}{{20}}} \right)\]
Simplifying the expression, we get
\[ \Rightarrow \] Required Probability \[ = {\left( {\dfrac{9}{{10}}} \right)^8}\left[ {{{\left( {\dfrac{9}{{10}}} \right)}^2} + 90 \times {{\left( {\dfrac{1}{{20}}} \right)}^2} + 20 \times \left( {\dfrac{9}{{10}}} \right) \times \left( {\dfrac{1}{{20}}} \right)} \right]\]
\[ \Rightarrow \] Required Probability \[ = {\left( {\dfrac{9}{{10}}} \right)^8}\left[ {\dfrac{{81}}{{100}} + \dfrac{{90}}{{400}} + \dfrac{{180}}{{200}}} \right]\]
Taking LCM inside the bracket, we get,
\[ \Rightarrow \] Required Probability \[ = {\left( {\dfrac{9}{{10}}} \right)^8}\left[ {\dfrac{{324 + 90 + 360}}{{400}}} \right]\]
Adding the terms, we get
\[ \Rightarrow \] Required Probability \[ = {\left( {\dfrac{9}{{10}}} \right)^8}\left[ {\dfrac{{387}}{{2 \times {{10}^2}}}} \right]\]
\[ \Rightarrow \] Required Probability \[ = 1 - \dfrac{{{9^8} \times 387}}{{2 \times {{10}^{10}}}}\] approx.
Simplifying the expression, we get
Therefore, the probability that the match is finished in 10 or less games \[ = 0.17\]
Hence, option C is the correct answer.

Note: While solving this question, we should know the difference between permutations and combinations. Permutation is an act of arranging the numbers, whereas combination is a method of selecting a group of numbers or elements in any order. Also, in order to answer this question, we should know that when we compute a factorial then, we write it in the form of: \[n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1\]. By factorial we mean that it is a product of all the positive integers which are less than or equal to the given number but not less than 1.