Entries of a $2 \times 2$ determinant are chosen from the set $\{-1, 1\}$. The probability that determinant has zero value is:
Step-by-Step Solution
Key Concept: Bayes' theorem relates conditional probabilities by accounting for both the likelihood of observing the event given each hypothesis and the prior probability of each hypothesis.
Using Bayes' theorem, we find $P(A_1|A)$ where $A_1$ is the event that the missing card is red and $A$ is drawing a red card from 51 remaining cards. Since there are 26 red and 26 black cards in a standard deck, $P(A_1) = \frac{1}{2}$ and $P(A_2) = \frac{1}{2}$. When the missing card is red, $P(A|A_1) = \frac{25}{51}$; when black, $P(A|A_2) = \frac{26}{51}$. Applying Bayes' formula: $P(A_1|A) = \frac{\frac{1}{2} \cdot \frac{25}{51}}{\frac{1}{2} \cdot \frac{25}{51} + \frac{1}{2} \cdot \frac{26}{51}} = \frac{25}{51}$.
Correct Answer: 3