Assertion (A): P(A' B) = 1 - P(A B). Reason (R): P(A B) + P(A' B) = P(S B) = 1, since A and A' are disjoint and make up S.

Conditional Probability & Bayes' Theorem · Question 15 of 125

Question 15

MediumOriginal question

Assertion (A): $P(A' \mid B) = 1 - P(A \mid B)$ .
Reason (R): $P(A \mid B) + P(A' \mid B) = P(S \mid B) = 1$ , since $A$ and $A'$ are disjoint and make up $S$ .

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 A

Correct answer: A

A is true for any $B$ with $P(B) > 0$ .
R is true and is exactly the reason: rearranging it gives A.

Concept: The complement rule under a condition
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsConditional probability

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Assertion (A): P(A′∣B)=1−P(A∣B)P(A' \mid B) = 1 - P(A \mid B)P(A′∣B)=1−P(A∣B).Reason (R): P(A∣B)+P(A′∣B)=P(S∣B)=1P(A \mid B) + P(A' \mid B) = P(S \mid B) = 1P(A∣B)+P(A′∣B)=P(S∣B)=1, since AAA and A′A'A′ are disjoint and make up SSS.

Explanation

Assertion (A): $P(A' \mid B) = 1 - P(A \mid B)$ .
Reason (R): $P(A \mid B) + P(A' \mid B) = P(S \mid B) = 1$ , since $A$ and $A'$ are disjoint and make up $S$ .