Question:

The probability that a student passes only in Hindi is 1/4. The probability of the student passes only in English is 5/9. The probability that the student passes in both of these subjects is 1/9. The probability that the student will pass in at least one of these two subjects is:

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Be careful with the wording "only." $P(\text{Hindi only})$ is $P(H) - P(H \cap E)$. Since "only" probabilities are already provided, you simply add them to the intersection probability to get the union.
Updated On: May 20, 2026
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The Correct Option is B

Solution and Explanation

Concept: The probability of "at least one" event occurring is the union of those events ($P(H \cup E)$). For two events, this is the sum of the probabilities of passing only $A$, only $B$, and both $A$ and $B$.

Step 1:
Identify given probabilities.

• $P(\text{Hindi only}) = 1/4$
• $P(\text{English only}) = 5/9$
• $P(\text{Hindi} \cap \text{English}) = 1/9$

Step 2:
Apply the addition rule for disjoint regions.
The event "passing at least one subject" is represented by: \[ P(H \cup E) = P(\text{Hindi only}) + P(\text{English only}) + P(\text{Hindi} \cap \text{English}) \] \[ P(H \cup E) = \frac{1}{4} + \frac{5}{9} + \frac{1}{9} \] \[ P(H \cup E) = \frac{1}{4} + \frac{6}{9} = \frac{1}{4} + \frac{2}{3} \]

Step 3:
Calculate the final sum.
Find the common denominator (12): \[ P(H \cup E) = \frac{3}{12} + \frac{8}{12} = \frac{11}{12} \]
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