A man is known to speak truth 7 out of 10 times. He threw a pair of dice and reports that doublet appeared. Find the probability that it was actually…

\(\displaystyle \left.\begin{array}{l}E_{1}: \text { doublet appeared } \\ E_{2}: \text { doublet did not appear } \\ A: \text { He reports doublet }\end{array}\right\}\) \[\begin{aligned} & P\left(E_{1}\right)=\frac{1}{6} \quad P\left(E_{2}\right)=\frac{5}{6} \\ & P\left(A / E_{1}\right)=\frac{7}{10} \quad P\left(A / E_{2}\right)=\frac{3}{10} \\ & P\left(E_{1} / A\right)=\frac{P\left(E_{1}\right) P\left(A / E_{1}\right)}{P\left(E_{1}\right) P\left(A / E_{1}\right)+P\left(E_{2}\right) P\left(A / E_{2}\right)} \\ & =\frac{\dfrac{1}{6} \cdot \dfrac{7}{10}}{\dfrac{1}{6} \cdot \dfrac{7}{10}+\dfrac{5}{6} \cdot \dfrac{3}{10}} \\ & =\frac{7}{22} \end{aligned} \]