Two unbiased dice are thrown. Then the probability that neither a doublet nor a total of 10 will appear is
Two unbiased dice are thrown. Then the probability that neither a doublet nor a total of 10 will appear is
$\frac{1}{12}$
$\frac{1}{36}$
$\frac{2}{9}$
$\frac{7}{9}$
Solution
Number of ways of getting doublet $=6$
Number of ways getting a total of 10 are $\Rightarrow(4,6),(5,5),(6,4)$ i.e. 3 ways
Here $(5,5)$ is common.
$\therefore$ Total ways of getting doublet or total of 10 are $6+3-1=8$
Hence required probability $=\frac{36-8}{36}=\frac{28}{36}=\frac{7}{9}$