Step 1: Define the events
Let \( E_1 \) be the event that the lost card is a King, and \( E_2 \) be the event that the lost card is not a King. Let \( A \) be the event of drawing a King from the remaining 51 cards.
Step 2: Assign probabilities to the events
\[ P(E_1) = \frac{1}{13}, \quad P(E_2) = \frac{12}{13}, \quad P(A|E_1) = \frac{3}{51}, \quad P(A|E_2) = \frac{4}{51} \]
Step 3: Use Bayes' Theorem
The required probability is \( P(E_1|A) \), which is given by: \[ P(E_1|A) = \frac{P(A|E_1) \cdot P(E_1)}{P(A|E_1) \cdot P(E_1) + P(A|E_2) \cdot P(E_2)} \] Substituting the values: \[ P(E_1|A) = \frac{\frac{1}{13} \cdot \frac{3}{51}}{\frac{1}{13} \cdot \frac{3}{51} + \frac{12}{13} \cdot \frac{4}{51}} = \frac{\frac{3}{663}}{\frac{3}{663} + \frac{48}{663}} = \frac{3}{51} = \frac{1}{17} \]
Step 4: Final result
The probability that the lost card is a King is \( \frac{1}{17} \).
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Four students of class XII are given a problem to solve independently. Their respective chances of solving the problem are: \[ \frac{1}{2},\quad \frac{1}{3},\quad \frac{2}{3},\quad \frac{1}{5} \] Find the probability that at most one of them will solve the problem.
The probability distribution of a random variable \( X \) is given below:
| \( X \) | 1 | 2 | 4 | 2k | 3k | 5k |
|---|---|---|---|---|---|---|
| \( P(X) \) | \( \frac{1}{2} \) | \( \frac{1}{5} \) | \( \frac{3}{25} \) | \( \frac{1}{10} \) | \( \frac{1}{25} \) | \( \frac{1}{25} \) |