Let E and F are two independent events such that P(E) = 0.35 and P(\( E\cup F \)) = 0.60 then which of following statements are TRUE?
(A) P(F) = \( \displaystyle\frac{5}{13} \)
(B) P(E|\( \overline{\mathrm{F}} \)) = 0.35
(C) P(E|\( \overline{\mathrm{F}} \)) = 0.65
(D) P(\( \overline{\mathrm{E}} \)|\( \overline{\mathrm{F}} \)) = 0.65
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
For independent events, \( P(E\cap F)=P(E)P(F) \). Let P(F) = p.
\( \displaystyle P(E\cup F)=P(E)+P(F)-P(E)P(F) \)
\( \displaystyle 0.60=0.35+p-0.35p \)
\( \displaystyle 0.25=0.65p \)
\( \displaystyle p=\frac{25}{65}=\frac{5}{13} \)
If E and F are independent, so are E and \( \overline{F} \), and so are \( \overline{E} \) and \( \overline{F} \). Hence conditioning on \( \overline{F} \) does not change the probabilities of E or \( \overline{E} \):
\( \displaystyle P(E|\overline{F})=P(E)=0.35 \)
\( \displaystyle P(\overline{E}|\overline{F})=P(\overline{E})=1-0.35=0.65 \)
So \( P(F)=\frac{5}{13} \), \( P(E|\overline{F})=0.35 \) and \( P(\overline{E}|\overline{F})=0.65 \).
Checking the statements
(A) P(F) = \( \frac{5}{13} \) — Correct. \( 0.25=0.65p \) gives \( p=\frac{5}{13} \).
(B) P(E|\( \overline{F} \)) = 0.35 — Correct. By independence, \( P(E|\overline{F})=P(E)=0.35 \).
(C) P(E|\( \overline{F} \)) = 0.65 — Incorrect. \( P(E|\overline{F}) \) is 0.35; 0.65 is \( P(\overline{E}) \).
(D) P(\( \overline{E} \)|\( \overline{F} \)) = 0.65 — Correct. \( P(\overline{E}|\overline{F})=P(\overline{E})=1-0.35=0.65 \).