Question

In a simultaneous throw of a pair of dice, if the probability of getting a sum less than $7$ is $\frac{a}{b}.$ Find $a+b.$

Answer 1

Let $S=$ Total outcomes

$=\left\{\right(1,1\left)\right(1,2\left)\right(1,3\left)\right(1,4\left)\right(1,5\left)\right(1,6)$

$(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)$

$(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)$

$(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)$

$(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)$

$(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\}$

$E=$ sum less than $7$

$=\left\{\right(1,1\left)\right(1,2\left)\right(1,3\left)\right(1,4\left)\right(1,5\left)\right(2,1\left)\right(2,2\left)\right(2,3\left)\right(2,4\left)\right(3,1\left)\right(3,2\left)\right(3,3\left)\right(4,1\left)\right(4,2\left)\right(5,1\left)\right\}$

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{15}{36}=\frac{5}{12}$

i.e probability of getting sum less than $7$ is $\frac{5}{12}=\frac{a}{b}$

$\Rightarrow a=5,b=12$

$\therefore a+b=5+12=17$