In a game two players <math><mi>A</mi></math> and <math><mi>B</mi></math> take turns in throwing a pair of fair dice starting with player <math><mi>A</mi></math> 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 <math><mi>B</mi></math> throws a total of 7 and <math><mi>B</mi></math> wins the game if he throws a total of 7 before <math><mi>A</mi></math> throws a total of six. The game stops as soon as either of the players wins. The probability of <math><mi>A</mi></math> winning the game is:

A2=cos?2θisin?2θisin?2θcos?2θSimilarly, A5=cos?5θisin?5θisin?5θcos?5θ=abcd(1) a2+b2=cos2?5θ-sin2?5θ=cos?10θ=cos?75?(2) a2-d2=cos2?5θ-cos2?5θ=0(3) a2-b2=cos2?5θ+sin2?5θ=1(4) a2-c2=cos2?5θ+sin2?5θ=1

In a game two players&nbsp;<math><mi>A</mi></math>&nbsp;and&nbsp;<math><mi>B</mi></math>&nbsp;take turns in throwing a pair of fair dice starting with player&nbsp;<math><mi>A</mi></math>&nbsp;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&nbsp;<math><mi>B</mi></math>&nbsp;throws a total of 7 and&nbsp;<math><mi>B</mi></math>&nbsp;wins the game if he throws a total of 7 before&nbsp;<math><mi>A</mi></math>&nbsp;throws a total of six. The game stops as soon as either of the players wins. The probability of&nbsp;<math><mi>A</mi></math>&nbsp;winning the game is:

<math> <mfrac> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> <mrow> <mrow> <mn>31</mn> </mrow> </mrow> </mfrac> </math> (Probability)

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths Class 12th Probability Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

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:

Option 1 -

$\frac{5}{6}$

Option 2 -

$\frac{31}{61}$

Option 3 -

$\frac{30}{61}$

Option 4 -

$\frac{5}{31}$ (Probability)

2 Views | Posted 2 months ago

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Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$A^{2} = (\begin{array}{l} c o s ? 2 θ & i s i n ? 2 θ \\ i s i n ? 2 θ & c o s ? 2 θ \end{array})$

Similarly, $A^{5} = (\begin{matrix} c o s ? 5 θ & i s i n ? 5 θ \\ i s i n ? 5 θ & c o s ? 5 θ \end{matrix}) = (\begin{array}{l} a & b \\ c & d \end{array})$

(1) $a^{2} + b^{2} = {c o s}^{2} ? 5 θ - {s i n}^{2} ? 5 θ = c o s ? 10 θ = c o s ? 75^{?}$

(2) $a^{2} - d^{2} = {c o s}^{2} ? 5 θ - {c o s}^{2} ? 5 θ = 0$

(3) $a^{2} - b^{2} = {c o s}^{2} ? 5 θ + {s i n}^{2} ? 5 θ = 1$

(4) $a^{2} - c^{2} = {c o s}^{2} ? 5 θ + {s i n}^{2} ? 5 θ = 1$

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