2.3 In an experiment involving a toxic substance, the probability that a white mouse will be alive for 10 hours is \( 7 / 10 \), and the probability that a black mouse will be alive for 10 hours is \( 9 / 10 \). Find the probability that, at the end of 10 hours, a) both mice will be alive b) only the black mouse will be alive c) only the white mouse will be alive d) at least one mouse will be alive

To solve this problem, we will denote the events as follows: - Let \( W \) be the event that the white mouse is alive after 10 hours. - Let \( B \) be the event that the black mouse is alive after 10 hours. From the problem statement, we have: - \( P(W) = \frac{7}{10} \) (the probability that the white mouse is alive) - \( P(B) = \frac{9}{10} \) (the probability that the black mouse is alive) We can also find the probabilities that each mouse is not alive after 10 hours: - \( P(W') = 1 - P(W) = 1 - \frac{7}{10} = \frac{3}{10} \) (the probability that the white mouse is not alive) - \( P(B') = 1 - P(B) = 1 - \frac{9}{10} = \frac{1}{10} \) (the probability that the black mouse is not alive) Assuming the events are independent, we can calculate the required probabilities: ### a) Probability that both mice will be alive \[ P(W \cap B) = P(W) \cdot P(B) = \frac{7}{10} \cdot \frac{9}{10} = \frac{63}{100} \] ### b) Probability that only the black mouse will be alive This occurs when the black mouse is alive and the white mouse is not alive: \[ P(B \cap W') = P(B) \cdot P(W') = \frac{9}{10} \cdot \frac{3}{10} = \frac{27}{100} \] ### c) Probability that only the white mouse will be alive This occurs when the white mouse is alive and the black mouse is not alive: \[ P(W \cap B') = P(W) \cdot P(B') = \frac{7}{10} \cdot \frac{1}{10} = \frac{7}{100} \] ### d) Probability that at least one mouse will be alive The probability that at least one mouse is alive can be found using the complement of the event that both mice are not alive: \[ P(\text{at least one alive}) = 1 - P(W' \cap B') = 1 - P(W') \cdot P(B') = 1 - \left(\frac{3}{10} \cdot \frac{1}{10}\right) = 1 - \frac{3}{100} = \frac{97}{100} \] ### Summary of Results - a) Probability that both mice will be alive: \( \frac{63}{100} \) - b) Probability that only the black mouse will be alive: \( \frac{27}{100} \) - c) Probability that only the white mouse will be alive: \( \frac{7}{100} \) - d) Probability that at least one mouse will be alive: \( \frac{97}{100} \)