Question

when two dice are thrown simulaneously, then the probability that the sum is neither 9 nor 12 is .

• A.
  $\frac{31}{36}$
• B.
  $\frac{8}{9}$
• C.
  $\frac{11}{12}$

Answer 1

The correct option is A

$\frac{31}{36}$
Let $A$ be the event of getting sum of digits as '9'
$B$ be the event of getting sum of digits as '12'

Favourable outcomes for $A$ are:
$\text{(3, 6), (4, 5), (5, 4), (6, 3)}$ $\Rightarrow 4$ number of outcomes.

∴ $\text{P (A)}$ $=\frac{4}{36}$ $=\frac{1}{9}$

Favourable outcomes for $B$ are ( 6, 6 ) $\Rightarrow 1$ outcome.

∴ $\text{P (B)}$ $=\frac{1}{36}$

Since A and B are mutually exclusive events, therefore
$\text{P(either A or B)}$$=\text{P(A)+ P(B)}$
$=\frac{1}{9}+\frac{1}{36}=\frac{5}{36}$
$P\left(\text{neither A nor B}\right)=1-\text{P(either 9 or 12)}$
$=1-\frac{5}{36}$$=\frac{31}{36}$