Question

Two dice were thrown simultaneously. Then find the probability of getting the sum of outcomes less than 7

• A. $\frac{7}{12}$
• B. $\frac{5}{12}$
• C. $\frac{5}{18}$
• D. $\frac{1}{9}$

Answer 1

The correct option is B $\frac{5}{12}$

The total possible outcomes will be $6\times 6=36$

They are :

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

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

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

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

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

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

Number of getting sum of outcomes less than 7 $=15$
[i.e.$(1,1)(1,2)(1,3)(1,4)(1,5)(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(4,1)(4,2)(5,1)$]

$P\left(A\right)=\frac{Numberoffavorableoutcomes}{TotalNumberofpossibleoutcomes}$
Probability $P\left(A\right)=\frac{15}{36}=\frac{5}{12}$

hence, option $B$ is correct.