Q15. In a school examination, the probabilities that David will pass Mathematics, English and Science are \( \frac{7}{8}, \frac{7}{10} \) and \( \frac{4}{5} \) respectively. Find the probability that David (a) passes English only. (b) passes English and Science only, (c) fails exactly two subjects, (d) passes at least one subject.

To solve the given problem, we first note the probabilities to pass each subject: - Probability David passes Mathematics, \( P(M) = \frac{7}{8} \) - Probability David passes English, \( P(E) = \frac{7}{10} \) - Probability David passes Science, \( P(S) = \frac{4}{5} \) Calculating the probabilities of failing each subject is straightforward: - Probability David fails Mathematics, \( P(M') = 1 - P(M) = \frac{1}{8} \) - Probability David fails English, \( P(E') = 1 - P(E) = \frac{3}{10} \) - Probability David fails Science, \( P(S') = 1 - P(S) = \frac{1}{5} \) Now, let's calculate each requirement: (a) To find the probability that David passes English only: This means he must pass English and fail both Mathematics and Science. \[ P(E \text{ only}) = P(E) \times P(M') \times P(S') = \frac{7}{10} \times \frac{1}{8} \times \frac{1}{5} = \frac{7}{400} \] (b) To find the probability that David passes both English and Science only: This requires that he passes English and Science but fails Mathematics. \[ P(E \text{ and } S \text{ only}) = P(E) \times P(S) \times P(M') = \frac{7}{10} \times \frac{4}{5} \times \frac{1}{8} = \frac{28}{400} = \frac{7}{100} \] (c) To find the probability that David fails exactly two subjects: He can fail Mathematics and English but pass Science, or he can fail Mathematics and Science but pass English, or he can fail English and Science but pass Mathematics. Calculating these: 1. Fails Mathematics and English, passes Science: \[ P(M' \cap E' \cap S) = P(M') \times P(E') \times P(S) = \frac{1}{8} \times \frac{3}{10} \times \frac{4}{5} = \frac{12}{400} = \frac{3}{100} \] 2. Fails Mathematics and Science, passes English: \[ P(M' \cap E \cap S') = P(M') \times P(E) \times P(S') = \frac{1}{8} \times \frac{7}{10} \times \frac{1}{5} = \frac{7}{400} \] 3. Fails English and Science, passes Mathematics: \[ P(M \cap E' \cap S') = P(M) \times P(E') \times P(S') = \frac{7}{8} \times \frac{3}{10} \times \frac{1}{5} = \frac{21}{400} \] Adding these probabilities gives: \[ P(\text{fails exactly 2}) = \frac{3}{100} + \frac{7}{400} + \frac{21}{400} = \frac{12}{400} + \frac{7}{400} + \frac{21}{400} = \frac{40}{400} = \frac{1}{10} \] (d) To find the probability that David passes at least one subject: Use the complementary method. The probability that he fails all subjects will be calculated first: \[ P(M') \times P(E') \times P(S') = \frac{1}{8} \times \frac{3}{10} \times \frac{1}{5} = \frac{3}{400} \] Then, the probability that he passes at least one subject is: \[ P(\text{at least one}) = 1 - P(\text{fails all}) = 1 - \frac{3}{400} = \frac{397}{400} \] Summarizing our answers: (a) \( \frac{7}{400} \) (b) \( \frac{7}{100} \) (c) \( \frac{1}{10} \) (d) \( \frac{397}{400} \)