Question

Let \(E\) and \(F\) be two events such that \[ P(\bar E)=\frac13,\qquad P(E\cap F)=\frac13,\qquad P(\overline{E\cup F})=\frac16. \] Which of the following is correct?

• A. \(E\) and \(F\) are not equally likely but independent
• B. \(E\) and \(F\) are mutually exclusive and independent
• C. \(E\) and \(F\) are equally likely but not independent
• D. \(E\) and \(F\) are independent and equally likely

Hint:

To check independence, verify \[ \boxed{P(E\cap F)=P(E)P(F).} \] To check whether two events are equally likely, compare \[ \boxed{P(E)\ \text{and}\ P(F).} \]

Answer 1

The Correct Option is A

Step 1: Find \(P(E)\) and \(P(E\cup F)\). Since \[ P(\bar E)=\frac13, \] we have \[ P(E)=1-\frac13=\frac23. \] Also, \[ P(\overline{E\cup F})=\frac16, \] so \[ P(E\cup F) = 1-\frac16 = \frac56. \]

Step 2: Find \(P(F)\). Using \[ P(E\cup F) = P(E)+P(F)-P(E\cap F), \] we get \[ \frac56 = \frac23+P(F)-\frac13. \] Hence, \[ P(F) = \frac12. \] Therefore, \[ P(E)\ne P(F), \] so the events are not equally likely.

Step 3: Check independence. Now, \[ P(E)P(F) = \frac23\times\frac12 = \frac13. \] Since \[ P(E\cap F)=\frac13=P(E)P(F), \] the events are independent. Therefore, \[ \boxed{\text{\(E\) and \(F\) are not equally likely but independent}.} \] Thus, \[ \boxed{(A)} \] is the correct answer.