Question

Two dice were thrown simultaneously. Then find the probability of getting at least one outcome as multiple of 3.

• A. $\frac{2}{9}$
• B. $\frac{1}{6}$
• C. $\frac{5}{12}$
• D. $\frac{5}{9}$

Answer 1

The correct option is D $\frac{5}{9}$

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 atleast one outcome as multiple of 3$=20$
[i.e.$(1,3)(1,6)(2,3)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,6)(5,3)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)$]

$P\left(A\right)=\frac{Numberoffavorableoutcomes}{TotalNumberofpossibleoutcomes}$
Probability $P\left(A\right)=\frac{20}{36}=\frac{5}{9}$

hence, option $D$ is correct.