Question

Two students appeared simultaneously for an entrance exam. If the probability that the first student gets qualified in the exam is \( \frac{1}{4} \) and the probability that the second student gets qualified in the same exam is \( \frac{2}{5} \), then the probability that at least one of them gets qualified in that exam is

• A. \( \frac{1}{10} \)
• B. \( \frac{7}{20} \)
• C. \( \frac{6}{10} \)
• D. \( \frac{11}{20} \)

Hint:

For independent events \( A \) and \( B \), the probability that at least one occurs is: \[ P(A \cup B) = P(A) + P(B) - P(A)P(B) \] Use this identity when you're asked for “at least one”.

Answer 1

The Correct Option is D

Step 1: Let the events be defined as follows
Let \( A \) be the event that the first student gets qualified. Let \( B \) be the event that the second student gets qualified. \[ P(A) = \frac{1}{4},
P(B) = \frac{2}{5} \] Step 2: Use the formula for probability of at least one event
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Assuming independence of events: \[ P(A \cap B) = P(A) . P(B) = \frac{1}{4} . \frac{2}{5} = \frac{2}{20} = \frac{1}{10} \] \[ P(A \cup B) = \frac{1}{4} + \frac{2}{5} - \frac{1}{10} \] Take LCM of 4, 5, and 10 which is 20: \[ = \frac{5}{20} + \frac{8}{20} - \frac{2}{20} = \frac{11}{20} \] Final Answer: \( \boxed{\frac{11}{20}} \)