Verified Solution • 2025–26 Session moderate

QJEEM20Ma3ec3: JEE Main 2020 Mathematics - Class 12 JEEM20Ma3ec3 Solution: In a game two players A and B take turns in throwing a pair of…

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 total a 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 :

Multiple Choice Options

  • A 56{5 \over {6}}
  • B 531{5 \over {31}}
  • C 3161{31 \over {61}}
  • D 3061{30 \over {61}}

Step-by-Step Verified Solution

Sum total 6 = {(1,5)(2,4)(3,3)(4,2)(5,1)}

P(6)=536P(6) = {5 \over 36}

Sum total 7 = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}

 P(7)=636\,P(7) = {6 \over {36}} = 16{1 \over 6}

Game ends and A wins if A throws 6 in 1st throw or A don’t throw 6 in 1st throw, B don’t throw 7 in 1st throw and then A throw 6 in his 2 nd chance and so on.

P(A) = A + A‾B‾A\overline A \overline B A + A‾B‾A‾B‾A\overline A \overline B \overline A \overline B A

= 536+(3136)(3036)(536)+{5 \over {36}} + \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)\left( {{5 \over {36}}} \right) + ..... ∞\infty

= 5361−(3136)(3036){{{5 \over {36}}} \over {1 - \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)}}

= 5×3636×36−31×30{{5 \times 36} \over {36 \times 36 - 31 \times 30}}

= 3061{30 \over {61}}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring