A student can solve 2 out of 4 problems of mathematics, 3 out of 5 problem of physics, and 4 out of 5 problems of chemistry. There are equal number of books of math, physics, and chemistry in his shelf. He selects one book randomly and attempts 10 problems from it. If he solves the first problem, then the probability that he will be able to solve the second problem is `2//3` b. `25//38` c. `13//21` d. `14//23`

Options

A. `2//3`
B. `25//38`
C. `13//21`
D. `14//23`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the probabilities of choosing each book
The student has an equal number of books for mathematics, physics, and chemistry. Therefore, the probability of choosing any one book is:

\[ P(M) = P(P) = P(C) = \frac{1}{3} \]

### Step 2: Calculate the probability of solving the first problem from each subject
- For Mathematics, the student can solve 2 out of 4 problems:
\[ P(S_1 | M) = \frac{2}{4} = \frac{1}{2} \]

- For Physics, the student can solve 3 out of 5 problems:
\[ P(S_1 | P) = \frac{3}{5} \]

- For Chemistry, the student can solve 4 out of 5 problems:
\[ P(S_1 | C) = \frac{4}{5} \]

### Step 3: Calculate the total probability of solving the first problem
Using the law of total probability:
\[ P(S_1) = P(M) \cdot P(S_1 | M) + P(P) \cdot P(S_1 | P) + P(C) \cdot P(S_1 | C) \]

Substituting the values:
\[ P(S_1) = \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{3}{5} + \frac{1}{3} \cdot \frac{4}{5} \]
\[ = \frac{1}{6} + \frac{1}{5} + \frac{4}{15} \]

Finding a common denominator (which is 30):
\[ = \frac{5}{30} + \frac{6}{30} + \frac{8}{30} = \frac{19}{30} \]

### Step 4: Calculate the probability of solving the second problem given the first problem was solved
- For Mathematics:
\[ P(S_2 | S_1, M) = \frac{1}{2} \]

- For Physics:
\[ P(S_2 | S_1, P) = \frac{3}{5} \]

- For Chemistry:
\[ P(S_2 | S_1, C) = \frac{4}{5} \]

Using the law of total probability again:
\[ P(S_2 | S_1) = P(M) \cdot P(S_2 | S_1, M) + P(P) \cdot P(S_2 | S_1, P) + P(C) \cdot P(S_2 | S_1, C) \]
\[ = \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{3}{5} + \frac{1}{3} \cdot \frac{4}{5} \]
\[ = \frac{1}{6} + \frac{1}{5} + \frac{4}{15} \]

Again, finding a common denominator (30):
\[ = \frac{5}{30} + \frac{6}{30} + \frac{8}{30} = \frac{19}{30} \]

### Step 5: Calculate the conditional probability
Now we need to find the probability that the second problem is solved given that the first problem was solved:
\[ P(S_2 | S_1) = \frac{P(S_2 \cap S_1)}{P(S_1)} \]
Where \(P(S_2 \cap S_1) = P(S_2 | S_1) \cdot P(S_1)\):
\[ P(S_2 | S_1) = \frac{P(S_2 | S_1)}{P(S_1)} = \frac{\frac{125}{300}}{\frac{19}{30}} = \frac{125}{300} \cdot \frac{30}{19} = \frac{25}{38} \]

### Final Answer
Thus, the probability that the student will be able to solve the second problem given that he solved the first problem is:
\[ \boxed{\frac{25}{38}} \]