The probability that a student passes at least in one of the three examinations A,B,C is 0.75, the probability that he passes in atleast two of the exams is 0.5 and the proabability he passes exactly two of the exams is 0.4 . If `alpha,beta, gamma`, are respectively the probabilities of the student passing in A,B,C then.

Options

A. `alpha+beta +gamma =(17)/(20)`
B. `alpha +beta +gamma=(27)/(20)`
C. `alphabetagamma=(1)/(10)`
D. `alpha^(2)+beta^(2)+gamma^(2)=1`.

Answer

To solve the problem step by step, we will use the information given about the probabilities of passing the exams A, B, and C.

### Step 1: Define the probabilities
Let:
- \( \alpha \) = Probability of passing exam A
- \( \beta \) = Probability of passing exam B
- \( \gamma \) = Probability of passing exam C

### Step 2: Use the first condition
The probability that a student passes at least one of the three examinations A, B, C is given as 0.75. This can be expressed using the complement rule:
\[ P(A \cup B \cup C) = 1 - P(A' \cap B' \cap C') = 0.75 \]
Thus,
\[ P(A' \cap B' \cap C') = 1 - 0.75 = 0.25 \]
Using independence, we can express this as:
\[ (1 - \alpha)(1 - \beta)(1 - \gamma) = 0.25 \]
This is our **Equation 1**.

### Step 3: Use the second condition
The probability that the student passes at least two of the exams is given as 0.5. This can be expressed as:
\[ P(\text{at least 2}) = P(\text{exactly 2}) + P(\text{all 3}) = 0.5 \]
From the problem, we know:
- \( P(\text{exactly 2}) = 0.4 \)

Let \( P(\text{all 3}) = P(A \cap B \cap C) = \alpha \beta \gamma \). Therefore:
\[ 0.4 + \alpha \beta \gamma = 0.5 \]
This simplifies to:
\[ \alpha \beta \gamma = 0.1 \]
This is our **Equation 2**.

### Step 4: Use the third condition
The probability of passing exactly two exams can be expressed as:
\[ P(\text{exactly 2}) = P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) \]
Using independence, this can be written as:
\[ \alpha \beta (1 - \gamma) + \alpha (1 - \beta) \gamma + (1 - \alpha) \beta \gamma = 0.4 \]
Expanding this gives:
\[ \alpha \beta - \alpha \beta \gamma + \alpha \gamma - \alpha \beta \gamma + \beta \gamma - \alpha \beta \gamma = 0.4 \]
Combining like terms:
\[ \alpha \beta + \alpha \gamma + \beta \gamma - 3\alpha \beta \gamma = 0.4 \]
This is our **Equation 3**.

### Step 5: Solve the equations
Now we have:
1. \( (1 - \alpha)(1 - \beta)(1 - \gamma) = 0.25 \) (Equation 1)
2. \( \alpha \beta \gamma = 0.1 \) (Equation 2)
3. \( \alpha \beta + \alpha \gamma + \beta \gamma - 3\alpha \beta \gamma = 0.4 \) (Equation 3)

Substituting \( \alpha \beta \gamma = 0.1 \) into Equation 3:
\[ \alpha \beta + \alpha \gamma + \beta \gamma - 0.3 = 0.4 \]
Thus,
\[ \alpha \beta + \alpha \gamma + \beta \gamma = 0.7 \]
This is our **Equation 4**.

### Step 6: Substitute and solve for individual variables
Now we can use Equations 1, 2, and 4 to find relationships between \( \alpha, \beta, \gamma \).

From Equation 1:
\[ (1 - \alpha)(1 - \beta)(1 - \gamma) = 0.25 \]
Expanding this gives:
\[ 1 - \alpha - \beta - \gamma + \alpha \beta + \beta \gamma + \gamma \alpha - \alpha \beta \gamma = 0.25 \]
Substituting \( \alpha \beta \gamma = 0.1 \) and \( \alpha \beta + \alpha \gamma + \beta \gamma = 0.7 \):
\[ 1 - (\alpha + \beta + \gamma) + 0.7 - 0.1 = 0.25 \]
This simplifies to:
\[ 1 - (\alpha + \beta + \gamma) + 0.6 = 0.25 \]
Thus,
\[ \alpha + \beta + \gamma = 1.35 \]

### Final Results
We have the following relationships:
1. \( \alpha \beta \gamma = 0.1 \)
2. \( \alpha + \beta + \gamma = 1.35 \)