A die is thrown twice and the stun of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Explanation

`=\frac{\cot \theta(\tan \theta-\tan 3 \theta)+\tan \theta(\cot \theta-\cot 3 \theta)}{(\cot \theta-\cot 3 \theta)(\tan \theta-\tan 3 \theta)}`


`=\frac{\cot \theta \tan \theta-\cot \theta \tan 3 \theta+\tan \theta \cot \theta-\tan \theta \cot 3 \theta}{\cot \theta \tan \theta-\cot \theta \tan 3 \theta-\cot 3 \theta \tan \theta+\cot 3 \theta \tan 3 \theta}`


`=\frac{1-\cot \theta \tan 3 \theta+1-\tan \theta \cot 3 \theta}{1-\cot \theta \tan 3 \theta-\cot 3 \theta \tan \theta+1}`


`=\frac{2-\cot \theta \tan 3 \theta-\tan \theta \cot 3 \theta}{2-\cot \theta \tan 3 \theta-\tan \theta \cot 3 \theta}=\frac{1}{\tan \theta} \Rightarrow \cot \theta \tan \theta=1`