To find the value of \(n\), we need to use the formula for the probability of the union of two independent events \(X\) and \(Y\):
\(P(X \cup Y) = P(X) + P(Y) - P(X)P(Y)\)
Given: \(P(X \cup Y) = 0.8\), \(P(X) = \frac{1}{3}\), \(P(Y) = n\).
Plugging the values into the formula:
\(0.8 = \frac{1}{3} + n - \left(\frac{1}{3}\right)n\)
To simplify, multiply the equation by 3 to eliminate the fractions:
\(3 \times 0.8 = 3 \times \frac{1}{3} + 3n - n\)
This results in:
\(2.4 = 1 + 2n\)
Subtract 1 from both sides:
\(1.4 = 2n\)
Divide by 2 to solve for \(n\):
\(n = \frac{1.4}{2}\)
\(n = \frac{7}{10}\)
Thus, the correct option is \(\frac{7}{10}\).
For independent events \( X \) and \( Y \), the probability of at least one occurring is given by:
\[ P(X \cup Y) = P(X) + P(Y) - P(X)P(Y). \]
Substituting the given values:
\[ 0.8 = \frac{1}{3} + n - \left(\frac{1}{3}\right)n. \]
Simplifying:
\[ 0.8 = \frac{1}{3} + n - \frac{n}{3}. \]
Combining terms:
\[ 0.8 = \frac{1}{3} + \frac{2n}{3}. \]
Multiplying the entire equation by 3:
\[ 2.4 = 1 + 2n. \]
Rearranging:
\[ 2n = 1.4 \implies n = 0.7 = \frac{7}{10}. \]
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 \( A_2 \)?