Step 1: Relationship between \(P(E)\) and \(P(\overline{E})\) We know that: \[ P(E) + P(\overline{E}) = 1. \]
Substituting \(P(\overline{E}) = 0.6\): \[ P(E) = 1 - 0.6 = 0.4. \]
Step 2: Use the formula for \(P(E \cup F)\) For any two events \(E\) and \(F\), we have: \[ P(E \cup F) = P(E) + P(F) - P(E \cap F). \] Since \(E\) and \(F\) are independent, \(P(E \cap F) = P(E) \cdot P(F)\).
Substituting this: \[ P(E \cup F) = P(E) + P(F) - P(E) \cdot P(F). \] Substitute \(P(E) = 0.4\) and \(P(E \cup F) = 0.6\): \[ 0.6 = 0.4 + P(F) - (0.4 \cdot P(F)). \] Simplify: \[ 0.6 = 0.4 + P(F) - 0.4P(F). \] \[ 0.6 - 0.4 = P(F)(1 - 0.4). \] \[ 0.2 = 0.6P(F). \] \[ P(F) = \frac{0.2}{0.6} = \frac{1}{3}. \]
Step 3: Find \(P(\overline{E} \cup \overline{F})\) Using the complement rule: \[ P(\overline{E} \cup \overline{F}) = 1 - P(E \cap F). \] From the formula for complements: \[ P(E \cap F) = P(E) \cdot P(F). \]
Substitute \(P(E) = 0.4\) and \(P(F) = \frac{1}{3}\): \[ P(E \cap F) = 0.4 \cdot \frac{1}{3} = \frac{2}{15}. \]
Thus: \[ P(\overline{E} \cup \overline{F}) = 1 - P(E \cap F) = 1 - \frac{2}{15} = \frac{15}{15} - \frac{2}{15} = \frac{13}{15}. \]
Final Answer: \[ P(F) = \frac{1}{3}, \quad P(\overline{E} \cup \overline{F}) = \frac{13}{15}. \]
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 \( A_2 \)?
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.