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.
There are 10 black and 5 white balls in a bag. Two balls are taken out, one after another, and the first ball is not placed back before the second is taken out. Assume that the drawing of each ball from the bag is equally likely. What is the probability that both the balls drawn are black?
The probabilities of solving a question by \( A \) and \( B \) independently are \( \frac{1}{2} \) and \( \frac{1}{3} \) respectively. If both of them try to solve it independently, find the probability that: