Giải ra hộ e với mn ơi

To solve this problem, we can use the concept of probability and the complement rule. The probability of exactly one person catching a fish can be calculated using the formula: \[P(\text{exactly one}) = P(A \cap \neg B \cap \neg C) + P(\neg A \cap B \cap \neg C) + P(\neg A \cap \neg B \cap C)\] where \(A\), \(B\), and \(C\) represent the events of the first, second, and third person catching a fish, respectively. Similarly, the probability of exactly two people catching a fish is given by: \[P(\text{exactly two}) = P(A \cap B \cap \neg C) + P(A \cap \neg B \cap C) + P(\neg A \cap B \cap C)\] The probability of at least one person catching a fish can be found using the complement rule: \[P(\text{at least one}) = 1 - P(\text{nobody catches a fish}) = 1 - P(\neg A \cap \neg B \cap \neg C)\] We can then substitute the given probabilities into these formulas to find the final answers. After calculation, we get: \[P(\text{exactly one}) = 0.5 * (1-0.4) * (1-0.3) + (1-0.5) * 0.4 * (1-0.3) + (1-0.5) * (1-0.4) * 0.3 = 0.44\] \[P(\text{exactly two}) = 0.5 * 0.4 * (1-0.3) + 0.5 * (1-0.4) * 0.3 + (1-0.5) * 0.4 * 0.3 = 0.29\] \[P(\text{at least one}) = 1 - (1-0.5)*(1-0.4)*(1-0.3) = 0 .79\] Therefore, the final answers are: a),"Xác suất để có đúng 1 người câu được cá bằng: $P(\text{exactly one})= 0 .44$", b),"Xác suất để có đúng 2 người câu được cá bằng: $P(\text{exactly two})= .29$", c),"Xác suất để người thứ ba luôn luôn câu được cá bằng: $P(C)= .3$", d),"Xác suất để có ít nhất một người câu được cá bằng: $P(at least one)= .79$"