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 \( \frac{1}{17} \).
List-I | List-II |
(A) Absolute maximum value | (I) 3 |
(B) Absolute minimum value | (II) 0 |
(C) Point of maxima | (III) -5 |
(D) Point of minima | (IV) 4 |
In number theory, it is often important to find factors of an integer \( N \). The number \( N \) has two trivial factors, namely 1 and \( N \). Any other factor, if it exists, is called a non-trivial factor of \( N \). Naresh has plotted a graph of some constraints (linear inequations) with points \( A(0, 50) \), \( B(20, 40) \), \( C(50, 100) \), \( D(0, 200) \), and \( E(100, 0) \). This graph is constructed using three non-trivial constraints and two trivial constraints. One of the non-trivial constraints is \( x + 2y \geq 100 \).
Based on the above information, answer the following questions:
On her birthday, Prema decides to donate some money to children of an orphanage home.
If there are 8 children less, everyone gets ₹ 10 more. However, if there are 16 children more, everyone gets ₹ 10 less. Let the number of children in the orphanage home be \( x \) and the amount to be donated to each child be \( y \).
Based on the above information, answer the following questions:
Let \( X \) denote the number of hours a Class 12 student studies during a randomly selected school day. The probability that \( X \) can take the values \( x_i \), for an unknown constant \( k \):
\[ P(X = x_i) = \begin{cases} 0.1, & {if } x_i = 0, \\ kx_i, & {if } x_i = 1 { or } 2, \\ k(5 - x_i), & {if } x_i = 3 { or } 4. \end{cases} \]The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: