
Two persons A and B take turns throwing a pair of dice. The first person to throw 9 from both dice will be awarded a prize. If A throws first then probability that B wins the game is
1. $\dfrac{9}{17}$
2. $\dfrac{8}{17}$
3. $\dfrac{8}{9}$
4. $\dfrac{1}{9}$
Answer
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Hint: First we will calculate the possibility of the total number of times 9 will come and then the possibility where nine will not come. Then, the probability of B winning and A losing considering the conditions given in the questions.
Complete step-by-step solution:
Nine can only come in these four ways (6, 3), (3, 6), (4, 5), (5, 4) where (number on the first dice, number on the second dice).
P (winning) = 4/36
= 1/9
P (not winning) = 1 – 1/9
= 8/9
B will win only when A throws the dice first. So, probability of B wins the game is
P (B winning) = $\left( {\dfrac{8}{9} \times \dfrac{1}{9} + \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{1}{9} + \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{1}{9}..........} \right)$
$ = \dfrac{1}{9}\left( {\dfrac{8}{9} + {{\left( {\dfrac{8}{9}} \right)}^3} + {{\left( {\dfrac{8}{9}} \right)}^5}.........} \right)$
This makes the equation in the bracket a GP.
Now, we will use the formula of sum of terms in GP.
Formula of the sum of the GP series is $\dfrac{{{a_1}}}{{1 - r}}$.
$a_1$ is the first digit of the series.
r is the common ratio.
$r = \left({\dfrac{8}{9}}\right)^2$
P (B winning) $=\dfrac{1}{9}.\left( {\dfrac{{\dfrac{8}{9}}}{{1 - {{\left( {\dfrac{8}{9}} \right)}^2}}}} \right)$
$\Rightarrow$ P (B winning) $ = \dfrac{1}{9}.\left( {\dfrac{{\dfrac{8}{9}}}{{1 - \dfrac{{64}}{{81}}}}} \right)$
$\Rightarrow$ P (B winning) $ = \dfrac{1}{9}.\left( {\dfrac{{\dfrac{8}{9}}}{{\dfrac{{81 - 64}}{{81}}}}} \right)$
$\Rightarrow$ P (B winning )$ = \dfrac{1}{9}.\left( {\dfrac{8}{9} \times \dfrac{{81}}{{17}}} \right)$
$\Rightarrow$ P (B winning) $ = \dfrac{8}{{17}}$
Probability of B winning the game is $\dfrac{8}{17}$.
So, option (2) is the correct answer.
Note: Probability is a possibility. We need to remember that in probability we use multiplication until the case is completed but we use addition when the case has been completed. Another thing to keep in mind is that each time we calculate the possibility it is necessary that A throws the dice first otherwise the answer will not be correct.
Complete step-by-step solution:
Nine can only come in these four ways (6, 3), (3, 6), (4, 5), (5, 4) where (number on the first dice, number on the second dice).
P (winning) = 4/36
= 1/9
P (not winning) = 1 – 1/9
= 8/9
B will win only when A throws the dice first. So, probability of B wins the game is
P (B winning) = $\left( {\dfrac{8}{9} \times \dfrac{1}{9} + \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{1}{9} + \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{8}{9} \times \dfrac{1}{9}..........} \right)$
$ = \dfrac{1}{9}\left( {\dfrac{8}{9} + {{\left( {\dfrac{8}{9}} \right)}^3} + {{\left( {\dfrac{8}{9}} \right)}^5}.........} \right)$
This makes the equation in the bracket a GP.
Now, we will use the formula of sum of terms in GP.
Formula of the sum of the GP series is $\dfrac{{{a_1}}}{{1 - r}}$.
$a_1$ is the first digit of the series.
r is the common ratio.
$r = \left({\dfrac{8}{9}}\right)^2$
P (B winning) $=\dfrac{1}{9}.\left( {\dfrac{{\dfrac{8}{9}}}{{1 - {{\left( {\dfrac{8}{9}} \right)}^2}}}} \right)$
$\Rightarrow$ P (B winning) $ = \dfrac{1}{9}.\left( {\dfrac{{\dfrac{8}{9}}}{{1 - \dfrac{{64}}{{81}}}}} \right)$
$\Rightarrow$ P (B winning) $ = \dfrac{1}{9}.\left( {\dfrac{{\dfrac{8}{9}}}{{\dfrac{{81 - 64}}{{81}}}}} \right)$
$\Rightarrow$ P (B winning )$ = \dfrac{1}{9}.\left( {\dfrac{8}{9} \times \dfrac{{81}}{{17}}} \right)$
$\Rightarrow$ P (B winning) $ = \dfrac{8}{{17}}$
Probability of B winning the game is $\dfrac{8}{17}$.
So, option (2) is the correct answer.
Note: Probability is a possibility. We need to remember that in probability we use multiplication until the case is completed but we use addition when the case has been completed. Another thing to keep in mind is that each time we calculate the possibility it is necessary that A throws the dice first otherwise the answer will not be correct.
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