Girl students constitute 10% of I year and 5% of II year at Roorkee University. During summer holidays 70% of the I year and 30% of II year students are given a project. The chance that I year girl student is on duty in a randomly selected day is

Options

A. `3/17`
B. `14/17`
C. `3/10`
D. `7/10`

Answer

To solve the problem step by step, we need to follow the given information and apply the concepts of probability.

### Step 1: Define the Variables
Let:
- \( G_1 \) = event that a student is a girl from the first year.
- \( G_2 \) = event that a student is a girl from the second year.
- \( P_1 \) = event that a student is given a project in the first year.
- \( P_2 \) = event that a student is given a project in the second year.

### Step 2: Calculate the Probabilities
From the problem, we know:
- The probability that a student is a girl in the first year, \( P(G_1) = 10\% = \frac{10}{100} = 0.1 \).
- The probability that a student is a girl in the second year, \( P(G_2) = 5\% = \frac{5}{100} = 0.05 \).
- The probability that a first-year student is given a project, \( P(P_1) = 70\% = \frac{70}{100} = 0.7 \).
- The probability that a second-year student is given a project, \( P(P_2) = 30\% = \frac{30}{100} = 0.3 \).

### Step 3: Calculate the Joint Probabilities
We need to find the probability that a randomly selected girl student is on duty. This can be calculated using the law of total probability.

1. The probability that a girl student from the first year is on duty:
\[ P(G_1 \cap P_1) = P(G_1) \times P(P_1) = 0.1 \times 0.7 = 0.07 \]

2. The probability that a girl student from the second year is on duty:
\[ P(G_2 \cap P_2) = P(G_2) \times P(P_2) = 0.05 \times 0.3 = 0.015 \]

### Step 4: Total Probability of a Girl Student Being on Duty
Now, we can find the total probability that a randomly selected student is a girl and is on duty:
\[ P(G \cap P) = P(G_1 \cap P_1) + P(G_2 \cap P_2) = 0.07 + 0.015 = 0.085 \]

### Step 5: Calculate the Total Probability of Being on Duty
Next, we need to find the total probability of being on duty, which includes both first-year and second-year students:
\[ P(P) = P(G_1 \cap P_1) + P(G_2 \cap P_2) + P(\text{not } G_1 \cap P_1) + P(\text{not } G_2 \cap P_2) \]

However, since we are only interested in the girls, we can directly use the probabilities we calculated.

### Step 6: Final Probability Calculation
Now, we can find the probability that a randomly selected student who is on duty is a girl:
\[ P(G | P) = \frac{P(G \cap P)}{P(P)} = \frac{0.085}{0.085 + (1 - 0.1) \times 0.7 + (1 - 0.05) \times 0.3} \]
Calculating the denominator:
- The probability of a boy from the first year on duty: \( 0.9 \times 0.7 = 0.63 \)
- The probability of a boy from the second year on duty: \( 0.95 \times 0.3 = 0.285 \)

Thus:
\[ P(P) = 0.085 + 0.63 + 0.285 = 1 \]
So:
\[ P(G | P) = \frac{0.085}{1} = 0.085 \]

### Conclusion
The chance that a first-year girl student is on duty on a randomly selected day is \( 0.085 \) or \( 8.5\% \).