Question

Two uniform dice marked $1$ to $6$ are thrown together The probability that the sum is neither $7$ nor $11$ is

• A. $\frac{2}{9}$
• B. $\frac{5}{9}$
• C. $\frac{7}{9}$
• D. $\frac{4}{9}$

Answer 1

Answer: (C)

$\frac{7}{9}$

The Sample Space will be
$S:$
$=(1,1),(1,2)...(1,6)$
$(2,1)...(2,6)$
:
:
:
$(6,1)...(6,6)$
Hence the total number of elements in the sample space is $6^2$
$=36$
Out of these there will be $8$ outcomes with sum $7$ or $11$.
Hence, total outcomes with sum neither $7$ or $11$ is
$36-8$
$=38$.
Hence the required probability is
$\frac{28}{36}$

$=\frac{7}{9}$