Question

In a simultaneous throw of a pair of dice, if the probability of getting neither $9$ nor $11$ as the sum of the numbers on the faces is $\frac{5}{a}.$ Find $a.$

Answer 1

The outcomes that we gets, in a simultaneous throw of a pair of dice are,

$S=\left\{\begin{array}{c}(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)\\ (2,1)(2,2)(2,3)(2,4)(2,5)(2,6)\\ (3,1)(3,2)(3,3)(3,4)(3,5)(3,6)\\ (4,1)(4,2)(4,3)(4,4)(4,5)(4,6)\\ (5,1)(5,2)(5,3)(5,4)(5,5)(5,6)\\ (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\end{array}\right\}$

$n\left(S\right)=36$

Now,

Favourable outcomes i.e getting neither $9$ nor $11$ as the sum of numbers on the faces are

$E=\left\{\begin{array}{c}(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)\\ (2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)\\ (3,4)(3,5)(4,1)(4,2)(4,3)(4,4)(4,6)\\ (5,1)(5,2)(5,3)(5,5)(6,1)(6,2)(6,4)(6,6)\end{array}\right\}$

$n\left(E\right)=30$

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{30}{36}$

$P\left(E\right)=\frac{5}{6}=\frac{5}{a}$

$\therefore a=6$