JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 4)
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 :
$${5 \over {6}}$$
$${5 \over {31}}$$
$${31 \over {61}}$$
$${30 \over {61}}$$
Explanation
Sum total 6 = {(1,5)(2,4)(3,3)(4,2)(5,1)}
$$P(6) = {5 \over 36}$$
Sum total 7 = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}
$$\,P(7) = {6 \over {36}}$$ = $${1 \over 6}$$
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 + $$\overline A \overline B A$$ + $$\overline A \overline B \overline A \overline B A$$
= $${5 \over {36}} + \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)\left( {{5 \over {36}}} \right) + $$ ..... $$\infty $$
= $${{{5 \over {36}}} \over {1 - \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)}}$$
= $${{5 \times 36} \over {36 \times 36 - 31 \times 30}}$$
= $${30 \over {61}}$$
$$P(6) = {5 \over 36}$$
Sum total 7 = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}
$$\,P(7) = {6 \over {36}}$$ = $${1 \over 6}$$
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 + $$\overline A \overline B A$$ + $$\overline A \overline B \overline A \overline B A$$
= $${5 \over {36}} + \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)\left( {{5 \over {36}}} \right) + $$ ..... $$\infty $$
= $${{{5 \over {36}}} \over {1 - \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)}}$$
= $${{5 \times 36} \over {36 \times 36 - 31 \times 30}}$$
= $${30 \over {61}}$$
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