Question

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

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

Answer 1

Answer: (C)

$\frac{575}{648}$

Let $E$ be the event that the throws of the two persons are unequal.
Then $E^{\prime }$ be the vent that the throws of the two persons are equal.
$\therefore$ The total number of cases for $E^{\prime }$ is ${\left(36\right)}^2$ $\Rightarrow n\left(S\right)={\left(36\right)}^2$
{ $\because S$ be the sample space }
We now proceed to find out the number of favorable cases for $E^{\prime }$.

Suppose ${(x+x^2+x^3+...+x^6)}^2=a_2{x}^2+a_3{x}^3+...+a_{12}{x}^{12}$
The number of favorable ways of $E^{\prime }={a_2}^2+{a_3}^2+...+{a_{12}}^2$
$\therefore n\left(E^{\prime }\right)=$ coefficient of constant term in $(a_2{x}^2+a_3{x}^3+...+a_{12}{x}^{12})\times (\frac{a_2}{x^2}+\frac{a_3}{x^3}+...+\frac{a_{12}}{x^{12}})$
$=$ coefficient of constant term in $\frac{{(1-x^6)}^2}{{(1-x)}^2}\times \frac{{(1-\frac{1}{x^6})}^2}{{(1-\frac{1}{x})}^2}$
$=$ coefficient of $x^{10}$ in ${(1-x^6)}^4{(1-x)}^{-4}$
$=$ coefficient of $x^{10}$ in $(1-4x^6+...)\left(1{+}^4{C}_{1}x{+}^5{C}_2{x}^2+...{+}^{13}{C}_{10}{x}^{10}+...\right)$

${=}^{13}{C}_{10}-{4.}^7{C}_4{=}^{13}{C}_3-{4.}^7{C}_3$$=\frac{\mathrm{13.12.11}}{\mathrm{1.2.3}}-4.\frac{\mathrm{7.6.5}}{\mathrm{1.2.3}}=146$

So the required probability is $=1-\frac{146}{1296}=1-\frac{73}{648}=\frac{575}{648}$