Question

$A$ and $B$ throw with $3$ dice; if $A$ throws sum of 8, $B$'s chance of throwing a higher number is $1-\frac{k}{27}$. Find the value of $k$

Answer 1

Here total number of ways $n=6^3=216.$
To find favourable number of ways, we have to find the sum of coefficients of powers of $x$ less than $9$ in the expansion.
${(x+x^2+x^3+x^4+x^5+x^6)}^3$ $=x^3{(1+x+x^2+x^3+x^4+x^5)}^3$ $=\frac{x^3{(1+x+x^2+x^3+x^4{x}^5)}^3{(1-x)}^3}{{(1-x)}^3}$ $=x^3{(1-x^6)}^3{(1-x)}^{-3}$ $=x^3[1-3x^6+3x^{12}-x^{16}]\times [1+3x+6x^2+.....+\frac{(r+1)(r+2)}{2}{x}^r+.....]$ $=[x^3-3x^9+3x^{15}-x^{21}]\times [1+3x+6x^2+....+\frac{(r+1)(r+2)}{2}{x}^r+.....]...\left(1\right)$
It is evident from (1) that $Coeff.of{x}^8=1\times \frac{(5+1)(5+2)}{2}=21$ $Coeff.of{x}^7=1\times \frac{(4+1)(4+2)}{2}=15$ $Coeff.of{x}^6=1\times \frac{(3+1)(3+2)}{2}=10$ $Coeff.of{x}^5=1\times \frac{(2+1)(2+2)}{2}=6$ $Coeff.of{x}^4=1\times \frac{(1+1)(1+2)}{2}=4$ and $Coeff.of{x}^3=1$
(Note that to obtain the coefficient of $x^3$ we substitute $r=5$ in the second bracket and multiply this coeff. of $x^5$ with the coefficient of $x^3$ (i.e.1) in the first bracket etc.)
Sum of these coeffcients$=56$
Hence the favourable number of ways. $m=216-56=160.$
Hence the required probability $=\frac{m}{n}=\frac{160}{216}=\frac{20}{27}.$