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 \( \frac{1}{17} \).
If the probability distribution is given by:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| P(x) | 0 | k | 2k | 2k | 3k | k² | 2k² | 7k² + k |
Then find: \( P(3 < x \leq 6) \)
If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :