Probability Calculation Steps

We are given the probabilities related to a student passing in Hindi and English:

• Probability of passing only in Hindi: P(H only) = $\frac{1}{4}$
• Probability of passing only in English: P(E only) = $\frac{5}{9}$
• Probability of passing in both Hindi and English: P(H $\cap$ E) = $\frac{1}{9}$

We need to find the probability that the student passes in at least one of these two subjects. This is represented by the union of the two events, P(H $\cup$ E).

The probability of passing in at least one subject can be calculated by summing the probabilities of the mutually exclusive events that make up the union:

P(H $\cup$ E) = P(H only) + P(E only) + P(H $\cap$ E)

Substitute the given values into the formula:

P(H $\cup$ E) = $\frac{1}{4} + \frac{5}{9} + \frac{1}{9}$

First, add the probabilities of passing only in English and passing in both:

$\frac{5}{9} + \frac{1}{9} = \frac{6}{9}$

Simplify the fraction $\frac{6}{9}$:

$\frac{6}{9} = \frac{2}{3}$

Now, add the probability of passing only in Hindi to this result:

P(H $\cup$ E) = $\frac{1}{4} + \frac{2}{3}$

To add these fractions, find a common denominator, which is 12:

• $\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$
• $\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}$

Add the fractions:

P(H $\cup$ E) = $\frac{3}{12} + \frac{8}{12} = \frac{11}{12}$

Therefore, the probability that the student will pass in at least one of these two subjects is $\frac{11}{12}$.