Question

Two persons each makes a single throw with a pair of dice. Then the probability that the sum on both dice are unequal, is:

• A. $\frac{57}{648}$
• B. $\frac{73}{648}$
• C. $\frac{575}{648}$
• D. $None$

Answer 1

The correct option is C $\frac{575}{648}$

Let number of ways for sum of number on two dice for both person will be $a_k$, where
$a_k$ =coefficent of $x^k$ in $(x+x^2+.......+x^6{)}^2.2\le k\le 12.$
$\Rightarrow (x+x^2+......+x^6{)}^2=a_2{x}^2+a_3{x}^2+a_3{x}^3+.....+a_{12}{x}^{12}$
Then
$a_2^2+a_3^2+........+a_{12}^2$= constant term in
$(a_2{x}^2+a_3{x}^3+.......+a_{12}{x}^{12}){(\frac{a_2}{x^2}+\frac{a_3}{x_3}+......+\frac{a_{12}}{x^{12}})}^2$
$=(x+x^2+......+x^6{)}^2{(\frac{1}{x}+\frac{1}{x^6}+.....+\frac{1}{x^6})}^6$
=Coefficient of $x^{10}$ in $(1+x+x^2+......+x^5{)}^4$
=Coefficient of $x^{10}$ in $(1-x^6{)}^4(1-x{)}^{-4}$
=Coefficient of $x^{10}$ in $(1-4x^6+......)(1-x{)}^{-4}$
${=}^{13}{C}_{10}-4{\times }^7{C}_4=146.$
Total ways = $(36{)}^2$
Hence probability (throws are unequal)
$=1-\frac{146}{36\times 36}=1-\frac{73}{648}=\frac{575}{648}$