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Two unbiased dice are thrown. Then the probability that neither a doublet nor a total of 10 will appear is
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The correct answer is:
$\frac{7}{9}$
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 $=\frac{36-8}{36}=\frac{28}{36}=\frac{7}{9}$
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 $=\frac{36-8}{36}=\frac{28}{36}=\frac{7}{9}$
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