There are two groups of subjects, one of which consists of 5 science subjects and 3 Engineering subjects and the other consists of 3 science and 5 Engineering subjects. An unbiased die is cast. If number 3 or 5 turns up, a subject from 1 is selected otherwise a subject is selected from group 2. The probability that an Engineering subject is selected ultimately, is

Options

A. `(7)/(13)`
B. `(9)/(17)`
C. `(13)/(24)`
D. `(11)/(20)`

Answer

To solve the problem, we need to find the probability that an Engineering subject is selected from either of the two groups based on the outcome of a die roll.

### Step-by-Step Solution:

1. **Identify the Groups:**
- Group 1: 5 Science subjects and 3 Engineering subjects (Total = 8 subjects)
- Group 2: 3 Science subjects and 5 Engineering subjects (Total = 8 subjects)

2. **Determine the Probability of Selecting Each Group:**
- The die is rolled, and if a 3 or 5 appears, we select from Group 1.
- The probability of rolling a 3 or 5 is:
\[ P(\text{Group 1}) = P(3 \text{ or } 5) = \frac{2}{6} = \frac{1}{3} \]
- If any other number (1, 2, 4, or 6) appears, we select from Group 2.
- The probability of rolling a number other than 3 or 5 is:
\[ P(\text{Group 2}) = P(1, 2, 4, \text{ or } 6) = \frac{4}{6} = \frac{2}{3} \]

3. **Calculate the Probability of Selecting an Engineering Subject from Each Group:**
- For Group 1, the probability of selecting an Engineering subject is:
\[ P(\text{Engineering | Group 1}) = \frac{\text{Number of Engineering subjects in Group 1}}{\text{Total subjects in Group 1}} = \frac{3}{8} \]
- For Group 2, the probability of selecting an Engineering subject is:
\[ P(\text{Engineering | Group 2}) = \frac{\text{Number of Engineering subjects in Group 2}}{\text{Total subjects in Group 2}} = \frac{5}{8} \]

4. **Use the Law of Total Probability:**
- The total probability of selecting an Engineering subject is given by:
\[ P(\text{Engineering}) = P(\text{Group 1}) \cdot P(\text{Engineering | Group 1}) + P(\text{Group 2}) \cdot P(\text{Engineering | Group 2}) \]
- Plugging in the values:
\[ P(\text{Engineering}) = \left(\frac{1}{3} \cdot \frac{3}{8}\right) + \left(\frac{2}{3} \cdot \frac{5}{8}\right) \]

5. **Calculate Each Term:**
- First term:
\[ \frac{1}{3} \cdot \frac{3}{8} = \frac{3}{24} \]
- Second term:
\[ \frac{2}{3} \cdot \frac{5}{8} = \frac{10}{24} \]

6. **Combine the Probabilities:**
- Now, add both terms:
\[ P(\text{Engineering}) = \frac{3}{24} + \frac{10}{24} = \frac{13}{24} \]

### Final Answer:
The probability that an Engineering subject is selected ultimately is:
\[ \frac{13}{24} \]