$A$ speaks truth $3$ times out of $4$, and $B$ $7$ times out of $10$, they both assert that a white ball has been drawn from a bag containing $6$ balls all of different colours; the probability of truth of the assertion is ______.
Step-by-Step Solution
Key Concept: Apply Bayes' theorem with two mutually exclusive hypotheses: H₁ (white ball actually drawn, probability 1/6) and H₂ (white ball not drawn, probability 5/6). Calculate P(white ball | both assert white) = P(both assert white | white) × P(white) / P(both assert white), where P(both assert white | white) = (3/4)(7/10) and P(both assert white | not white) = (1/4)(3/10).
There are two hypotheses: (i) their coincident testimony is true, (ii) it is false. With $p_1 = \frac{1}{6}$, $p_2 = \frac{5}{6}$, we have $p_1 p_2 = \frac{3}{4} \cdot \frac{1}{10} = \frac{3}{40}$ and $p_2' = \frac{1}{25} \cdot \frac{1}{4} \cdot \frac{3}{10} = \frac{1}{25}$ (accounting for the chance both draw the white ball when not drawn: $\frac{1}{5} \times \frac{1}{5} = \frac{1}{25}$). The ratio of probabilities is $35:1$, giving probability $\frac{35}{36}$.
Correct Answer: 0.9722