Assertion (A): If P(E) = 0.3, P(F) = 0.4 and P(E F) = 0.7, then E and F are not independent. Reason (R): If P(E F) = P(E) + P(F), then E and F are independent.

Question 70

MediumOriginal question

Assertion (A): If $P(E) = 0.3$ , $P(F) = 0.4$ and $P(E \cup F) = 0.7$ , then $E$ and $F$ are not independent.
Reason (R): If $P(E \cup F) = P(E) + P(F)$ , then $E$ and $F$ are independent.

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$P(E \cap F) = 0.3 + 0.4 - 0.7 = 0$ , while $P(E)\,P(F) = 0.12$ : A is true.
R is false: $P(E \cup F) = P(E) + P(F)$ means $P(E \cap F) = 0$ , so the events are mutually exclusive, and with positive probabilities that rules out independence.

Worth noting

Exclusive events with positive probabilities are never independent: if one happens, the other cannot.

Concept: Independence and exclusiveness
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsIndependent events

Practise this chapter →

Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.

Assertion (A): If P(E)=0.3P(E) = 0.3P(E)=0.3, P(F)=0.4P(F) = 0.4P(F)=0.4 and P(E∪F)=0.7P(E \cup F) = 0.7P(E∪F)=0.7, then EEE and FFF are not independent.Reason (R): If P(E∪F)=P(E)+P(F)P(E \cup F) = P(E) + P(F)P(E∪F)=P(E)+P(F), then EEE and FFF are independent.

Explanation

Assertion (A): If $P(E) = 0.3$ , $P(F) = 0.4$ and $P(E \cup F) = 0.7$ , then $E$ and $F$ are not independent.
Reason (R): If $P(E \cup F) = P(E) + P(F)$ , then $E$ and $F$ are independent.