Question

Aman and Bhuvan throw a pair of dice alternatively. In order to win, they have to get a sum of $ 8. $ Find their respective probabilities of winning if Aman starts the game.

Answer 1

Hint: Probability is the state of being probable and the extent to which something is likely to happen in the particular situations or the favourable outcomes. Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes.
 $ P(A) = \dfrac{\text{Total number of the favourable outcomes}}{\text{Total number of the outcomes}} $

Complete step-by-step answer:
Given that the two dice are thrown.
Outcomes will be –
\[
  (1,1)\;{\text{(1,2) (1,3) (1,4) (1,5) (1,6)}} \\
  (2,1)\;{\text{(2,2) (2,3) (2,4) (2,5) (2,6)}} \\
  (3,1)\;{\text{(3,2) (3,3) (3,4) (3,5) (3,6)}} \\
  (4,1)\;{\text{(4,2) (4,3) (4,4) (4,5) (4,6)}} \\
  (5,1)\;{\text{(5,2) (5,3) (5,4) (5,5) (5,6)}} \\
  (6,1)\;{\text{(6,2) (6,3) (6,4) (6,5) (6,6)}} \;
 \]
The total possible outcomes from the two thrown dice will be $ = 36 $ .....(A)
The favourable outcomes that is sum of the two dices should be $ = 8 $
Favourable outcomes =
 $ (2,6){\text{ (6,2) (4,4) (5,3) (3,5)}} $
The total number of favourable outcomes $ = 5 $ .... (B)
The probability that A wins –
 $ P(A) = \dfrac{5}{{36}} $ .... (I)
The probability that A fails is –
 $ P(\overline A ) = 1 - P(A) $
Place the values in the above equation –
 $ P(\overline A ) = 1 - \dfrac{5}{{36}} $
 $
   \Rightarrow P(A) = \dfrac{{36 - 5}}{{36}} \\
   \Rightarrow P(A) = \dfrac{{31}}{{36}}\;{\text{ }}....{\text{ (II)}} \\
  $
Similarly for The probability that B wins –
 $ P(B) = \dfrac{5}{{36}} $ .... (III)
The probability that B fails is –
 $ P(\overline B ) = 1 - P(B) $
Place the values in the above equation –
 $ P(\overline B ) = 1 - \dfrac{5}{{36}} $
 $
   \Rightarrow P(B) = \dfrac{{36 - 5}}{{36}} \\
   \Rightarrow P(B) = \dfrac{{31}}{{36}}\;{\text{ }}....{\text{ (IV)}} \\
  $
Hence, the probability that A wins –
 $ = \dfrac{5}{{36}} + \dfrac{{31}}{{36}} \times \dfrac{{31}}{{36}} \times \dfrac{5}{{36}} + \dfrac{{31}}{{36}} \times \dfrac{{31}}{{36}} \times \dfrac{{31}}{{36}} \times \dfrac{{31}}{{36}} \times \dfrac{5}{{36}} + ..... $
The above equation can be re-written as –
 $ = \dfrac{5}{{36}} + {\left( {\dfrac{{31}}{{36}}} \right)^2} \times \dfrac{5}{{36}} + {\left( {\dfrac{{31}}{{36}}} \right)^4} \times \dfrac{5}{{36}} + ..... $
The above series follows the Geometric progression, simplification implies –
 $ = \dfrac{{\dfrac{5}{{36}}}}{{1 - {{\left( {\dfrac{{31}}{{36}}} \right)}^2}}} $
Simplify the above equation –
 $ = \dfrac{{\dfrac{5}{{36}}}}{{\dfrac{{{{36}^2} - {{31}^2}}}{{{{36}^2}}}}} $
 $ = \dfrac{5}{{36}} \times \dfrac{{{{36}^2}}}{{335}} $
Common multiple from the numerator and the denominator cancel each other.
  $ = \dfrac{{36}}{{67}} $ is the required answer.
So, the correct answer is “ $ \dfrac{{36}}{{67}} $ ”.

Note: While simplification identify the sequence followed and apply formula accordingly, whether it is arithmetic progression or the geometric progression. The probability of any event always ranges between zero and one. It can never be the negative number or the number greater than one. The probability of impossible events is always equal to zero whereas, the probability of the sure event is always equal to one