Question

A and B throw a pair of dice alternately. A wins the game if he gets a total of 9 and B wins the game if he gets a total of 7. If A starts the game, then find the probability of winning the game by B.

Answer 1

Hint: Firstly find the probability that A wins and find the probability that B wins and then find the probability of winning the game by B the probability is given by:
 P(A LOSE)P(B WIN) + P(A LOSE)P(B LOSE)P(A LOSE)P(B WIN) +. . . . . . . . . . .
The probability of the complement of an event is found by subtracting the probability of an event from one.

Complete step-by-step solution:
Given A and B throw a pair of dice alternately. The sample space for the throw of a pair of a dice is given as.
\[\left\{ \begin{align}
  & \left( 1,1 \right),\left( 1,2 \right),\left( 1,3 \right),\left( 1,4 \right),\left( 1,5 \right),\left( 1,6 \right) \\
 & \left( 2,1 \right),\left( 2,2 \right),\left( 2,3 \right),\left( 2,4 \right),\left( 2,5 \right),\left( 2,6 \right) \\
 & \left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right) \\
 & \left( 4,1 \right),\left( 4,2 \right),\left( 4,3 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right) \\
 & \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 5,6 \right) \\
 & \left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,5 \right),\left( 6,6 \right) \\
\end{align} \right\}\]
Given A wins the game if he gets a total of 9
The possible outcomes for A winning the game is
\[\left\{ \begin{align}
  & \left( 3,6 \right),\left( 6,3 \right) \\
 & \left( 4,5 \right),\left( 5,4 \right) \\
\end{align} \right\}\]
The total number of possible outcomes when dice is thrown twice is 36.
The probability of A winning the game is equal to:
 \[P\left( \text{A WIN} \right)=\dfrac{4}{36}=\dfrac{1}{9}\].
Then the probability of A losing the game is calculated by subtracting 1 from P(A WIN).
$\begin{align}
  & P\left( \text{A LOSE} \right)=1-\dfrac{1}{9} \\
 & \Rightarrow P\left( \text{A LOSE} \right)=\dfrac{9-1}{9}=\dfrac{8}{9} \\
\end{align}$
The possible outcomes for B winning the game is
\[\left\{ \begin{align}
  & \left( 1,6 \right),\left( 6,1 \right) \\
 & \left( 2,5 \right),\left( 5,2 \right) \\
 & \left( 3,4 \right),\left( 4,3 \right) \\
\end{align} \right\}\]
The total number of possible outcomes when dice is thrown twice is 36.
The probability of B winning the game is equal to:
 \[P(\text{B WIN})=\dfrac{6}{36}=\dfrac{1}{6}\]
Now, the probability of B losing the game is equal to:
$\begin{align}
  & P\left( \text{B LOSE} \right)=1-P\left( \text{B WIN} \right) \\
 & \Rightarrow P\left( \text{B LOSE} \right)=1-\dfrac{1}{6} \\
 & \Rightarrow P\left( \text{B LOSE} \right)=\dfrac{5}{6} \\
\end{align}$
We have to find the probability of winning the game by B is
= P(A LOSE)P(B WIN) + P(A LOSE)P(B LOSE)P(A LOSE)P(B WIN) +. . . . . . . . . . .
\[=\dfrac{8}{9}\times \dfrac{1}{6}+\dfrac{8}{9}\times \dfrac{5}{6}\times \dfrac{8}{9}\times \dfrac{1}{6}+....\]
We have an infinite sum of series formed above, the sum of infinite series is given by relation \[S=\dfrac{a}{1-r}\], where a is the first term and r is the common ratio.
The common ratio is the term by which the infinite series progresses. To find the common ratio divide the first two terms of the infinite series.
\[\begin{align}
&r=\dfrac{\dfrac{8}{9}\times\dfrac{5}{6}\times\dfrac{8}{9}\times\dfrac{1}{6}}{\dfrac{8}{9}\times \dfrac{1}{6}} \\
& =\dfrac{5}{6}\times \dfrac{8}{9} \\
& =\dfrac{40}{54}
\end{align}\]
Therefore, the probability of B winning the game such that A starts is given by.
\[\begin{align}
  & P\left( B \right)=\dfrac{\dfrac{8}{9}\times \dfrac{1}{6}}{1-\dfrac{40}{54}} \\
 & =\dfrac{8}{54-40} \\
 & =\dfrac{8}{14} \\
 & =\dfrac{4}{7}
\end{align}\]

Note: The mistake that could be possible is in writing the sample space. You might have missed one or many possibilities. The sample space that we have written above is as follows:
\[\left\{ \begin{align}
  & \left( 1,1 \right),\left( 1,2 \right),\left( 1,3 \right),\left( 1,4 \right),\left( 1,5 \right),\left( 1,6 \right) \\
 & \left( 2,1 \right),\left( 2,2 \right),\left( 2,3 \right),\left( 2,4 \right),\left( 2,5 \right),\left( 2,6 \right) \\
 & \left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 3,4 \right),\left( 3,5 \right),\left( 3,6 \right) \\
 & \left( 4,1 \right),\left( 4,2 \right),\left( 4,3 \right),\left( 4,4 \right),\left( 4,5 \right),\left( 4,6 \right) \\
 & \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right),\left( 5,5 \right),\left( 5,6 \right) \\
 & \left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right),\left( 6,5 \right),\left( 6,6 \right) \\
\end{align} \right\}\]
The trick to write this sample space in such a way that you won’t make any mistakes is when writing the first row of the sample space, write 1 with 1 then 1 with 2 then 1 with 3 and so on till 1 with 6. Here, we are going till 6 because in a dice there are only 6 possibilities from 1 to 6. Similarly, we can write the third row, 2 with 1, 2 with 2, 2 with 3 and so on till 2 with 6. In this way, we can write all the rows of the sample space.