To find \( P(A \cup B) \), we can use the formula of probability for the union of two events: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
First, we need to find \( P(A \cap B) \), which is the intersection of events \( A \) and \( B \).
We use the conditional probability formula: \( P(A \cap B) = P(A | B) \cdot P(B) \).
Given \( P(A | B) = 0.6 \) and \( P(B) = 0.8 \), we compute \( P(A \cap B) \) as follows:
\[ P(A \cap B) = 0.6 \times 0.8 = 0.48 \]
Now we can find \( P(A \cup B) \):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = 0.4 + 0.8 - 0.48 = 0.72 \]
Therefore, the correct answer is 0.72.
The formula for the union of two events is:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
The conditional probability \( P(A \mid B) \) is related to \( P(A \cap B) \) as:
\( P(A \cap B) = P(A \mid B) \cdot P(B) \).
Substitute the given values:
\( P(A \cap B) = (0.6)(0.8) = 0.48 \).
Now calculate \( P(A \cup B) \):
\( P(A \cup B) = 0.4 + 0.8 - 0.48 = 0.72 \).
Thus, the probability \( P(A \cup B) \) is 0.72.
Based upon the results of regular medical check-ups in a hospital, it was found that out of 1000 people, 700 were very healthy, 200 maintained average health and 100 had a poor health record.
Let \( A_1 \): People with good health,
\( A_2 \): People with average health,
and \( A_3 \): People with poor health.
During a pandemic, the data expressed that the chances of people contracting the disease from category \( A_1, A_2 \) and \( A_3 \) are 25%, 35% and 50%, respectively.
Based upon the above information, answer the following questions:
(i) A person was tested randomly. What is the probability that he/she has contracted the disease?}
(ii) Given that the person has not contracted the disease, what is the probability that the person is from category \( A_2 \)?