A student takes examination in four subjects A, B, C, D. His chances of passing in A are `4/5`, in B are `3/4`, in C are `5/6` and in D are `2/3`. To qualify he must pass in A and atleast in two other subjects. What is the probability that he qualifies ?

Answer

To solve the problem step by step, we need to find the probability that the student qualifies by passing in subject A and at least two other subjects (B, C, or D).

### Step 1: Define the probabilities of passing each subject.
Let:
- \( P(A) = \frac{4}{5} \) (Probability of passing in A)
- \( P(B) = \frac{3}{4} \) (Probability of passing in B)
- \( P(C) = \frac{5}{6} \) (Probability of passing in C)
- \( P(D) = \frac{2}{3} \) (Probability of passing in D)

### Step 2: Define the probabilities of failing each subject.
The probabilities of failing each subject are:
- \( P(A') = 1 - P(A) = 1 - \frac{4}{5} = \frac{1}{5} \)
- \( P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4} \)
- \( P(C') = 1 - P(C) = 1 - \frac{5}{6} = \frac{1}{6} \)
- \( P(D') = 1 - P(D) = 1 - \frac{2}{3} = \frac{1}{3} \)

### Step 3: Identify the qualifying combinations.
The student qualifies if he passes A and at least two of the other subjects (B, C, D). The possible combinations are:
1. Pass A, B, C, D
2. Pass A, B, C, and fail D
3. Pass A, B, and D, and fail C
4. Pass A, C, and D, and fail B

### Step 4: Calculate the probabilities for each combination.
1. **Pass A, B, C, D:**
\[ P(A \cap B \cap C \cap D) = P(A) \cdot P(B) \cdot P(C) \cdot P(D) = \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{2}{3} \]
\[ = \frac{4 \cdot 3 \cdot 5 \cdot 2}{5 \cdot 4 \cdot 6 \cdot 3} = \frac{120}{360} = \frac{1}{3} \]

2. **Pass A, B, C, and fail D:**
\[ P(A \cap B \cap C \cap D') = P(A) \cdot P(B) \cdot P(C) \cdot P(D') = \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{1}{3} \]
\[ = \frac{4 \cdot 3 \cdot 5 \cdot 1}{5 \cdot 4 \cdot 6 \cdot 3} = \frac{60}{360} = \frac{1}{6} \]

3. **Pass A, B, and D, and fail C:**
\[ P(A \cap B \cap D \cap C') = P(A) \cdot P(B) \cdot P(D) \cdot P(C') = \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{6} \]
\[ = \frac{4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 6} = \frac{24}{360} = \frac{1}{15} \]

4. **Pass A, C, and D, and fail B:**
\[ P(A \cap C \cap D \cap B') = P(A) \cdot P(C) \cdot P(D) \cdot P(B') = \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{2}{3} \cdot \frac{1}{4} \]
\[ = \frac{4 \cdot 5 \cdot 2 \cdot 1}{5 \cdot 6 \cdot 3 \cdot 4} = \frac{40}{360} = \frac{1}{9} \]

### Step 5: Add the probabilities of all qualifying combinations.
Now, we sum the probabilities of all qualifying combinations:
\[ P(\text{Qualify}) = P(A \cap B \cap C \cap D) + P(A \cap B \cap C \cap D') + P(A \cap B \cap D \cap C') + P(A \cap C \cap D \cap B') \]
\[ = \frac{1}{3} + \frac{1}{6} + \frac{1}{15} + \frac{1}{9} \]

### Step 6: Find a common denominator and calculate the total probability.
The least common multiple of 3, 6, 15, and 9 is 90. Converting each fraction:
\[ = \frac{30}{90} + \frac{15}{90} + \frac{6}{90} + \frac{10}{90} = \frac{61}{90} \]

### Final Answer
Thus, the probability that the student qualifies is:
\[ \boxed{\frac{61}{90}} \]