Question

The probability that a student passes only in Mathematics is \( \frac{1}{3} \). The probability that the student passes only in English is \( \frac{4}{9} \). The probability that the student passes in both of these subjects is \( \frac{1}{6} \). The probability that the student will pass in at least one of these two subjects is

• A. \( \frac{17}{18} \)
• B. \( \frac{11}{18} \)
• C. \( \frac{14}{18} \)
• D. \( \frac{1}{18} \)

Hint:

When solving probability problems, ensure to consider mutual exclusivity or overlapping events and use the inclusion-exclusion principle accordingly.

Answer 1

The Correct Option is A

Step 1: Given probabilities are: \[ P(\text{only Mathematics}) = \frac{1}{3}, \quad P(\text{only English}) = \frac{4}{9}, \quad P(\text{both subjects}) = \frac{1}{6} \] Step 2: The probability of passing in at least one of the subjects is calculated using the principle of inclusion-exclusion: \[ P(\text{at least one}) = P(\text{only Mathematics}) + P(\text{only English}) + P(\text{both subjects}) \] Step 3: Substituting the values: \[ \frac{1}{3} + \frac{4}{9} + \frac{1}{6} \] Converting all fractions to a common denominator of 18: \[ \frac{6}{18} + \frac{8}{18} + \frac{3}{18} = \frac{17}{18} \] Conclusion: The probability that the student will pass in at least one of the subjects is \( \frac{17}{18} \), which corresponds to option (A).