A and B are any to events such that `P(A)=0.3, P(B)=0.8 and P(A cap B) =0.16`, find the probability that exactly one of the events happens.

Answer

To find the probability that exactly one of the events A or B occurs, we can follow these steps:

### Step 1: Identify the probabilities given in the problem.
We have:
- \( P(A) = 0.3 \)
- \( P(B) = 0.8 \)
- \( P(A \cap B) = 0.16 \)

### Step 2: Calculate the probability of only event A occurring.
The probability of only event A occurring can be calculated using the formula:
\[ P(\text{only A}) = P(A) - P(A \cap B) \]
Substituting the values:
\[ P(\text{only A}) = 0.3 - 0.16 = 0.14 \]

### Step 3: Calculate the probability of only event B occurring.
Similarly, the probability of only event B occurring can be calculated as:
\[ P(\text{only B}) = P(B) - P(A \cap B) \]
Substituting the values:
\[ P(\text{only B}) = 0.8 - 0.16 = 0.64 \]

### Step 4: Find the probability that exactly one of the events occurs.
The probability that exactly one of the events occurs is the sum of the probabilities of only A and only B:
\[ P(\text{exactly one of A or B}) = P(\text{only A}) + P(\text{only B}) \]
Substituting the values:
\[ P(\text{exactly one of A or B}) = 0.14 + 0.64 = 0.78 \]

### Step 5: Convert the decimal to a fraction.
To express \( 0.78 \) as a fraction:
\[ 0.78 = \frac{78}{100} \]
Now simplifying:
\[ \frac{78}{100} = \frac{39}{50} \]

### Final Answer:
The probability that exactly one of the events happens is \( \frac{39}{50} \).

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