To solve the problem, we need to find the probability \(P(A|B)\), where event \(A\) is that the first ball is black, and event \(B\) is that the second ball is black. The probability formula for conditional probability is:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Step 1: Find \(P(A \cap B)\).
The probability that the first ball is black and the second ball is also black can be calculated by considering the following:
Therefore:
\[ P(A \cap B) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \]
Step 2: Find \(P(B)\).
\(P(B)\) is the probability that the second ball is black regardless of the color of the first ball. Consider the two scenarios:
Thus:
\[ P(B) = \left(\frac{6}{10} \times \frac{5}{9}\right) + \left(\frac{4}{10} \times \frac{6}{9}\right) = \frac{30}{90} + \frac{24}{90} = \frac{54}{90} = \frac{3}{5} \]
Step 3: Calculate \(P(A|B)\).
Using the conditional probability formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9} \]
With gcd(5, 9) = 1, the fraction \(\frac{m}{n}\) is in its simplest form with \(m = 5\) and \(n = 9\). Hence, \(m + n\) equals \(14\).
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 \( A_2 \)?