Question

It has been found that if A and B play a game 12 times, A wins 6 times, B wins 4 times, and they draw twice. A and B take part in a series of 3 games. The probability that they win alternately, is:

Answer 1

Hint: In the given question, we have been given that there are some given number of players. They play some given number of games and the outcomes of the different players winning and drawing has been given. Then it has been given that the players play some given number of games. We have to find the probability of a certain event occurring in a certain pattern. We are going to solve it by first calculating the number of ways the certain pattern is possible, then calculate the separate probability of each pattern, add them up and find the answer.

Complete step by step solution:
Let the probability of A winning be \[p\].
Then, \[p = \dfrac{6}{{12}} = \dfrac{1}{2}\]
Let the probability of B winning be \[q\].
Then, \[q = \dfrac{4}{{12}} = \dfrac{1}{3}\]
Now, it has been given that A and B play three games, and we have to calculate the probability that they will win alternately.
There are two ways that this arrangement is possible – ABA or BAB.
Hence, \[p\left( {ABA} \right) = p \times q \times p = {p^2}q = {\left( {\dfrac{1}{2}} \right)^2}\left( {\dfrac{1}{3}} \right) = \dfrac{1}{{12}}\]
And \[p\left( {BAB} \right) = q \times p \times q = {q^2}p = {\left( {\dfrac{1}{3}} \right)^2}\left( {\dfrac{1}{2}} \right) = \dfrac{1}{{18}}\]
Since they are independent events, we can add them to find the probability, so,
\[P\left( {alt.} \right) = \dfrac{1}{{12}} + \dfrac{1}{{18}} = \dfrac{5}{{36}}\]

Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we think about the concept or formula which contains the known and the unknown, consider the different cases and calculate them separately. For solving such questions, we only need to know the definition of the main, the pivotal terms which are responsible for the answer to the given question.