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:
Step 1: Define probabilities. Let \( P(A) = \frac{1}{2} \) be the probability that \( A \) solves the problem. Let \( P(B) = \frac{1}{3} \) be the probability that \( B \) solves the problem.
Step 2: Compute probability that neither solves it. \[ P(\text{none}) = P(\text{A fails}) \times P(\text{B fails}) \] Since the events are independent, \[ P(A'B') = (1 - P(A)) \times (1 - P(B)) \] \[ = \left( 1 - \frac{1}{2} \right) \times \left( 1 - \frac{1}{3} \right) \] \[ = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \] Thus, the probability that none of them solve it is \( \frac{1}{3} \).
Step 3: Compute probability that at least one solves it. Using the complement rule: \[ P(\text{at least one}) = 1 - P(\text{none}) \] \[ = 1 - \frac{1}{3} = \frac{2}{3} \] Thus, the probability that at least one of them solves it is \( \frac{2}{3} \).
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 given events \( A \) and \( B \) are such that \( P(A) = \frac{1}{4} \), \( P(B) = \frac{1}{2} \), and \( P(A \cap B) = \frac{1}{8} \); then find \( P(A' \cap B') \).