Question

What is the probability of getting at least $1$ when a die is thrown twice.

• A. $\frac{11}{36}$
• B. $\frac{25}{36}$
• C. $\frac{15}{36}$
• D. $\frac{13}{36}$

Answer 1

Answer: (A)

$\frac{11}{36}$

Given: Two dice thrown

To find the probability of getting at least 1

Two dice thrown, so $n\left(S\right)=6\times 6=36$

Favourable outcomes of getting atleast $1$ is

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

$\Rightarrow n\left(E\right)=11$

Hence, the probability of getting at least $1$ is $\frac{n\left(E\right)}{n\left(S\right)}=\frac{11}{36}$