Question

In a single throw of two dice, what is the probability that neither a doublet nor a total of 8 will appear?

• A. $\frac{11}{36}$
• B. $\frac{5}{18}$
• C. $\frac{13}{18}$
• D. $\frac{3}{16}$

Answer 1

The correct option is B $\frac{5}{18}$

$n\left(S\right)=36$

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

$B=(2,6),(3,5),(4,4),(5,3),(6,2)$

$n\left(A\right)=6,n\left(B\right)=5,n(A\cap B)=1$

∴Required probability =$P(A\cup B)$ $=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=\frac{6}{36}+\frac{5}{36}-\frac{1}{36}=\frac{5}{18}$