\[ P(A) = \frac{1}{10}, \quad P(B|A) = \frac{3}{5}, \quad P(A|B^c) = \frac{1}{5} \] Using the law of total probability: \[ P(B) = P(B|A) P(A) + P(B|A^c) P(A^c) \] Since, \[ P(A^c) = 1 - P(A) = 1 - \frac{1}{10} = \frac{9}{10} \] And given: \[ P(B|A^c) = 1 - P(A|B^c) = 1 - \frac{1}{5} = \frac{4}{5} \] Substituting the values: \[ P(B) = \frac{3}{5} \times \frac{1}{10} + \frac{4}{5} \times \frac{9}{10} \] \[ = \frac{3}{50} + \frac{36}{50} = \frac{39}{50} = 0.78 \]