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14.Probability
hard
In a game two players $A$ and $B$ take turns in throwing a pair of fair dice starting with player $A$ and total of scores on the two dice, in each throw is noted. $A$ wins the game if he throws a total of $6$ before $B$ throws a total of $7$ and $B$ wins the game if he throws a total of $7$ before $A$ throws a total of six The game stops as soon as either of the players wins. The probability of $A$ winning the game is
A
$\frac{31}{61}$
B
$\frac{5}{6}$
C
$\frac{5}{31}$
D
$\frac{30}{61}$
(JEE MAIN-2020)
Solution
$P(6)=\frac{1}{6}, P(7)=\frac{5}{36}$
$P(A)=W+F F W+F F F F W+\ldots . .$
$=\frac{1}{6}+\frac{5}{6} \times \frac{31}{36} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{2}\left(\frac{31}{36}\right)^{2} \frac{1}{6}+\ldots$
$=\frac{\frac{1}{6}}{1-\frac{155}{216}}=\frac{36}{61}$
Standard 11
Mathematics
