A die is thrown twice. Let us represent the event 'obtaining an odd number on the first throw' by A and the event 'obtaining an odd number on the second throw' by B. Test the independency of the events A and B.
Step 1: The events \( A \) and \( B \) are independent if: \[ P(A \cap B) = P(A) \cdot P(B). \]
Step 2: For a fair die, the probability of rolling an odd number (1, 3, or 5) is \( \frac{3}{6} = \frac{1}{2} \), so: \[ P(A) = P(B) = \frac{1}{2}. \]
Step 3: The probability of both events occurring (i.e., rolling an odd number on both throws) is: \[ P(A \cap B) = P(\text{odd on first throw and odd on second throw}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. \]
Step 4: Now check if \( P(A \cap B) = P(A) \cdot P(B) \): \[ P(A \cap B) = \frac{1}{4}, \quad P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. \] Since they are equal, the events \( A \) and \( B \) are independent.
Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, where x>y, be 8 and 16 respectively. Two numbers are chosen from \(\{1, 2, 3, x-4, y, 5\}\) one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is:
If the mean and the variance of the data 
are $\mu$ and 19 respectively, then the value of $\lambda + \mu$ is