When a student is selected from a class consisting of 25 girls and a certain number of boys, the probability of the student being a boy is 3/8 . If 2 students were selected from this class, what would be the probability of exactly one girl and one boy being selected?

Options

A. 25/52
B. 23/52
C. 27/52
D. 21/52

Answer

To solve the problem step by step, we need to find the probability of selecting exactly one girl and one boy from a class consisting of 25 girls and a certain number of boys, given that the probability of selecting a boy is \( \frac{3}{8} \).

### Step 1: Determine the number of boys in the class
Let the number of boys be \( x \). The total number of students in the class is \( 25 + x \). The probability of selecting a boy is given by:

\[ P(\text{Boy}) = \frac{x}{25 + x} = \frac{3}{8} \]

### Step 2: Set up the equation
From the probability equation, we can cross-multiply to eliminate the fraction:

\[ 8x = 3(25 + x) \]

### Step 3: Simplify the equation
Expanding the right side gives:

\[ 8x = 75 + 3x \]

Now, subtract \( 3x \) from both sides:

\[ 8x - 3x = 75 \]

This simplifies to:

\[ 5x = 75 \]

### Step 4: Solve for \( x \)
Dividing both sides by 5 gives:

\[ x = 15 \]

So, there are 15 boys in the class.

### Step 5: Calculate the total number of students
The total number of students is:

\[ 25 + 15 = 40 \]

### Step 6: Calculate the probability of selecting exactly one girl and one boy
We can have two scenarios for selecting one girl and one boy:
1. The first student is a girl and the second is a boy.
2. The first student is a boy and the second is a girl.

#### Scenario 1: First girl, then boy
- Probability of selecting a girl first:

\[ P(\text{Girl first}) = \frac{25}{40} \]

- Probability of selecting a boy second (after one girl has been selected):

\[ P(\text{Boy second}) = \frac{15}{39} \]

So, the combined probability for this scenario is:

\[ P(\text{Girl first, Boy second}) = \frac{25}{40} \times \frac{15}{39} \]

#### Scenario 2: First boy, then girl
- Probability of selecting a boy first:

\[ P(\text{Boy first}) = \frac{15}{40} \]

- Probability of selecting a girl second (after one boy has been selected):

\[ P(\text{Girl second}) = \frac{25}{39} \]

So, the combined probability for this scenario is:

\[ P(\text{Boy first, Girl second}) = \frac{15}{40} \times \frac{25}{39} \]

### Step 7: Add the probabilities of both scenarios
Now we can add the probabilities from both scenarios:

\[ P(\text{Exactly one girl and one boy}) = P(\text{Girl first, Boy second}) + P(\text{Boy first, Girl second}) \]

Calculating this gives:

\[ P = \left(\frac{25}{40} \times \frac{15}{39}\right) + \left(\frac{15}{40} \times \frac{25}{39}\right) \]

This simplifies to:

\[ P = 2 \times \left(\frac{25}{40} \times \frac{15}{39}\right) = \frac{2 \times 25 \times 15}{40 \times 39} \]

### Step 8: Simplify the final probability
Calculating the numerator and denominator:

\[ = \frac{750}{1560} \]

Now, simplifying \( \frac{750}{1560} \):

Both numbers can be divided by 30:

\[ = \frac{25}{52} \]

### Final Answer
Thus, the probability of selecting exactly one girl and one boy is:

\[ \frac{25}{52} \]