Question

A and B play a match in which the chances of their winning a game is \[\dfrac{1}{4}\] for both of them and \[\dfrac{1}{2}\] is the probability that match is being drawn. The match is finished as soon as either player wins two games. Find the probability that the match will be finished in 4 or fewer games.
\[\left( a \right)\dfrac{9}{16}\]
\[\left( b \right)\dfrac{7}{16}\]
\[\left( c \right)\dfrac{1}{2}\]
\[\left( d \right)\dfrac{1}{4}\]

Answer 1

Hint: To solve this question, we will calculate the probability of the ways such that the match is not finished in 4 games. We will make all the possible points such that the match is not finished in less than 4 games and add them up to get all. Finally, we will use \[P\left( \overline{A} \right)=1-P\left( A \right)\] or \[P\left( A \right)=1-P\left( \overline{A} \right)\] to calculate the required probability from the obtained probability.

Complete step by step answer:
We have been given that the chances of A winning the game is \[\dfrac{1}{4}.\] And also the chances of B winning the game is \[\dfrac{1}{4}.\] The probability of draw is \[\dfrac{1}{2}\] . We have to find the probability that the match will be finished in 4 or fewer games. We will find ways in which the match is not finished in 4 games. To make that happen, there are 4 not related or mutually exclusive ways possible.
(1) When all 4 games are drawn.
The probability of that will be \[{{\left( \dfrac{1}{2} \right)}^{4}}\] as \[\dfrac{1}{2}\] is the probability of a match to be drawn and we are having 4 matches.
(2) A and B both win one game and the remaining two matches are drawn.
We will get that its probability is given by
\[^{4}{{C}_{2}}{{\left( \dfrac{1}{2} \right)}^{2}}{{\left( \dfrac{1}{4} \right)}^{2}}\]
As chances of winning of A and B are,
\[\Rightarrow \left( \dfrac{1}{4} \right)\times \left( \dfrac{1}{4} \right)={{\left( \dfrac{1}{4} \right)}^{2}}\]
And of two matches drawn is \[{{\left( \dfrac{1}{2} \right)}^{2}}.\] And also we have to select 2 from 4, so \[^{4}{{C}_{2}}\] is taken.
(3) A wins one game and the remaining 3 are drawn.
We will get that its probability is given by
\[^{4}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{3}}\left( \dfrac{1}{4} \right)\]
Thus drawn will be \[{{\left( \dfrac{1}{2} \right)}^{3}}.\]
\[\text{A win}=\left( \dfrac{1}{4} \right)\]
And 1 is selected from 4, so \[^{4}{{C}_{1}}\] is used.
(4) Lastly B wins the game and the remaining 3 are drawn. Its probability is given by
\[^{4}{{C}_{1}}\left( \dfrac{1}{4} \right){{\left( \dfrac{1}{2} \right)}^{3}}\]
(Again same as (3))
The total probability that the match is not finished in 4 games is the sum of all (1), (2), (3) and (4) and it is given by,
\[\text{Required Probability}={{\left( \dfrac{1}{2} \right)}^{4}}+{{\text{ }}^{4}}{{C}_{2}}{{\left( \dfrac{1}{2} \right)}^{2}}{{\left( \dfrac{1}{4} \right)}^{2}}+{{\text{ }}^{4}}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{3}}\left( \dfrac{1}{4} \right)+{{\text{ }}^{4}}{{C}_{1}}\left( \dfrac{1}{4} \right){{\left( \dfrac{1}{2} \right)}^{3}}\]
\[\Rightarrow \text{Required Probability}=\dfrac{1}{16}+\dfrac{3}{16}+\dfrac{1}{8}+\dfrac{1}{8}\]
\[\Rightarrow \text{Required Probability}=\dfrac{1}{2}\]
Now, we know that,
\[P\left( A \right)=1-P\left( \overline{A} \right)\]
where A is any event.
Therefore, we get,
\[P\left( \text{the match finished 4 or less than 4 games} \right)=1-\dfrac{1}{2}\]
\[\Rightarrow P\left( \text{the match finished 4 or less than 4 games} \right)=\dfrac{1}{2}\]
Therefore, \[\dfrac{1}{2}\] is the required probability.
Hence, the correct option is (c).

Note:
 The key point to note here is that (3) and (4) are almost the same because \[P\left( A\text{ winning} \right)=P\left( B\text{ winning} \right)=\dfrac{1}{4}\] as they are the same therefore, we could also have used \[2\times \left( 3 \right)\] in case of (3) and (4). Both ways are correct and give the same answer. Also, another key point to note here is that we have not used the formula, \[\text{Probability}=\dfrac{\text{Favorable Outcomes}}{\text{Total number of outcomes}}.\] This is because all the information given in the question was already in probability, so we need not apply the formula at last.