Question

Mr. A has two fair cubic dice one with faces numbered from 2 to 7 and the second with faces numbered from 4 to 9 . Twice, he randomly picks one of the dice (each dice is equally likely) and rolls. If it is known that the sum of the resulting two rolls is 10 then the probability he rolled the same dice twice is $\frac{m}{n}(m,n\in N)$, find the least value of $(n-m)$.

Answer 1

Answer: 5

Let $E_1:1^{\text{st }}$ dice selected twice
$E_1=\{(3,7)(7,3)(4,6)(6,4)(5,5)\}P\left(E_1\right)=\frac{5}{36}$
Let $E_2:2^{\text{nd }}$ dice selected twice
$E_2=\{(4,6)(6,4)(5,5)\}P\left(E_2\right)=\frac{3}{36}$
$E_3:1^{\text{st }}$ drawn $1^{\text{st }}$ dice and $2^{\text{nd }}$ drawn $2^{\text{nd }}$ dice
$E_3=\{(2,8),(3,7),(4,6),(5,5),(6,4)\}P\left(E_3\right)=\frac{5}{36}$
$E_4:1^{\text{st }}$ draw $2^{\text{nd }}$ dice and $2^{\text{nd }}$ draw $1^{\text{st }}$ dice
$E_4:\{(8,2)(7,3)(6,4)(5,5)(4,6)\}P\left(E_4\right)=\frac{5}{36}$
$P$ (same dice rolled twice/ sum is 10 )
$=\frac{\frac{1}{2}\times \frac{1}{2}\left(\frac{5}{36}+\frac{3}{36}\right)}{\frac{1}{2}\times \frac{1}{2}\left(\frac{5}{36}+\frac{3}{36}+\frac{5}{36}+\frac{5}{36}\right)}=\frac{8}{18}=\frac{4}{9}$