Consider two events \(T\) and \(S\). Let \(\overline{T}\) denote the complement of the event \(T\). The probability associated with different events are given as follows:
\[ P(\overline{T}) = 0.6, \qquad P(S|T) = 0.3, \qquad P(S|\overline{T}) = 0.6 \]
Then, \(P(T|S)\) is ______ (rounded off to two decimal places).
Answer
0.25
Explanation
First, \(P(T) = 1 - P(\overline{T}) = 0.4\). By the law of total probability,
\[ P(S) = P(S|T)P(T) + P(S|\overline{T})P(\overline{T}) = 0.3 \times 0.4 + 0.6 \times 0.6 = 0.12 + 0.36 = 0.48 \]
Bayes' theorem then reverses the conditioning:
\[ P(T|S) = \frac{P(S|T)P(T)}{P(S)} = \frac{0.12}{0.48} = 0.25 \]
Note that \(P(T|S) = 0.25\) differs from the given \(P(S|T) = 0.3\); confusing the two is the common trap.
Answer: 0.25