The probabilities of three mutually exclusive events A,B,C are : `P(A)=2//3,P(B)=1//4,P(C)=1//6`. Is the statement

Options

A. 1
B. wrong
C. could be either
D. do not know

Answer

To determine whether the statement regarding the probabilities of the mutually exclusive events A, B, and C is correct, we will follow these steps:

### Step 1: Identify the given probabilities
We have the following probabilities:
- \( P(A) = \frac{2}{3} \)
- \( P(B) = \frac{1}{4} \)
- \( P(C) = \frac{1}{6} \)

### Step 2: Check the definition of mutually exclusive events
Mutually exclusive events are events that cannot occur at the same time. Therefore, the intersection of any two events is zero:
- \( P(A \cap B) = 0 \)
- \( P(B \cap C) = 0 \)
- \( P(C \cap A) = 0 \)

### Step 3: Calculate the probability of the union of the events
The probability of the union of mutually exclusive events is the sum of their individual probabilities:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) \]

### Step 4: Substitute the values and find a common denominator
To add the probabilities, we need a common denominator. The least common multiple (LCM) of the denominators 3, 4, and 6 is 12.

Now, convert each probability:
- \( P(A) = \frac{2}{3} = \frac{8}{12} \)
- \( P(B) = \frac{1}{4} = \frac{3}{12} \)
- \( P(C) = \frac{1}{6} = \frac{2}{12} \)

### Step 5: Add the converted probabilities
Now, we can add these fractions:
\[ P(A \cup B \cup C) = \frac{8}{12} + \frac{3}{12} + \frac{2}{12} = \frac{13}{12} \]

### Step 6: Analyze the result
The result \( \frac{13}{12} \) is greater than 1. Since the probability of any event must be between 0 and 1, this indicates that the statement is incorrect.

### Conclusion
The statement regarding the probabilities of the mutually exclusive events A, B, and C is wrong because the total probability exceeds 1.

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