If \( A \) and \( B \) are two events such that \( P(\overline{A}\cup\overline{B})=\displaystyle\frac{2}{3} \) and \( P(A\cup B)=\displaystyle\frac{5}{9} \), then which of the following are TRUE?
(A) \( P(A\cap B)=\displaystyle\frac{1}{3} \)
(B) \( P(A\cup B)+P(A\cap B)=\displaystyle\frac{7}{9} \)
(C) \( P(A)+P(B)=\displaystyle\frac{2}{9} \)
(D) \( P(\overline{A})+P(\overline{B})=\displaystyle\frac{10}{9} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (D) only
By De Morgan's law, \( \overline{A}\cup\overline{B}=\overline{A\cap B} \), so
\( \displaystyle P(A\cap B)=1-P(\overline{A}\cup\overline{B}) \)
\( \displaystyle =1-\frac{2}{3}=\frac{1}{3} \)
By the addition rule, \( P(A)+P(B)=P(A\cup B)+P(A\cap B) \):
\( \displaystyle P(A)+P(B)=\frac{5}{9}+\frac{1}{3}=\frac{8}{9} \)
Then
\( \displaystyle P(\overline{A})+P(\overline{B})=2-\left[P(A)+P(B)\right] \)
\( \displaystyle =2-\frac{8}{9}=\frac{10}{9} \)
So \( P(A\cap B)=\frac{1}{3} \), \( P(A\cup B)+P(A\cap B)=\frac{8}{9} \), \( P(A)+P(B)=\frac{8}{9} \) and \( P(\overline{A})+P(\overline{B})=\frac{10}{9} \).
Checking the statements
(A) \( P(A\cap B)=\frac{1}{3} \) — Correct. \( P(A\cap B)=1-\frac{2}{3}=\frac{1}{3} \).
(B) \( P(A\cup B)+P(A\cap B)=\frac{7}{9} \) — Incorrect. The sum is \( \frac{5}{9}+\frac{3}{9}=\frac{8}{9} \).
(C) \( P(A)+P(B)=\frac{2}{9} \) — Incorrect. \( P(A)+P(B)=\frac{8}{9} \).
(D) \( P(\overline{A})+P(\overline{B})=\frac{10}{9} \) — Correct. \( 2-\frac{8}{9}=\frac{10}{9} \).
Tip: \( P(\overline{A}\cup\overline{B}) \) is the complement of \( P(A\cap B) \), not of \( P(A\cup B) \); mixing these up is the usual slip in this type.