Two thirds of eth students in a class are boys and the rest girls. It is known that eh probability of a girl getting a first class is 0.25 and that of a boy getting a first class is 0.28. find the probability that a student chosen at random will get first class marks in the subject.

Explanation

To solve the problem step-by-step, we will use the law of total probability.

### Step 1: Define Events
Let:
- \( E_1 \): The event that a selected student is a boy.
- \( E_2 \): The event that a selected student is a girl.
- \( A \): The event that a selected student gets first class marks.

### Step 2: Determine Probabilities of Events
From the problem:
- The probability of selecting a boy, \( P(E_1) = \frac{2}{3} \).
- The probability of selecting a girl, \( P(E_2) = 1 - P(E_1) = 1 - \frac{2}{3} = \frac{1}{3} \).

### Step 3: Determine Conditional Probabilities
- The probability of a boy getting first class marks, \( P(A | E_1) = 0.28 \).
- The probability of a girl getting first class marks, \( P(A | E_2) = 0.25 \).

### Step 4: Apply the Law of Total Probability
Using the law of total probability, we can find the probability that a randomly selected student gets first class marks:
\[ P(A) = P(E_1) \cdot P(A | E_1) + P(E_2) \cdot P(A | E_2) \]

### Step 5: Substitute Values
Substituting the known values into the equation:
\[ P(A) = \left( \frac{2}{3} \cdot 0.28 \right) + \left( \frac{1}{3} \cdot 0.25 \right) \]

### Step 6: Calculate Each Term
Calculating the first term:
\[ \frac{2}{3} \cdot 0.28 = \frac{0.56}{3} \]
Calculating the second term:
\[ \frac{1}{3} \cdot 0.25 = \frac{0.25}{3} \]

### Step 7: Combine the Results
Now, we can combine the results:
\[ P(A) = \frac{0.56}{3} + \frac{0.25}{3} = \frac{0.56 + 0.25}{3} = \frac{0.81}{3} \]

### Step 8: Final Calculation
Calculating \( \frac{0.81}{3} \):
\[ P(A) = 0.27 \]

### Conclusion
The probability that a student chosen at random will get first class marks in the subject is \( 0.27 \).

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