P(A) = 1/3, P(B) = 1/4 and P(A B) = 1/12. What is P(A B'), the probability of A but not B?

Question 70

HardOriginal question

$P(A) = \frac{1}{3}$ , $P(B) = \frac{1}{4}$ and $P(A \cap B) = \frac{1}{12}$ . What is $P(A \cap B')$ , the probability of $A$ but not $B$ ?

\frac{1}{12}

Correct answerYour answer

\frac{1}{6}

Correct answerYour answer

\frac{1}{4}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$A$ splits into the part inside $B$ and the part outside it: $P(A) = P(A \cap B) + P(A \cap B')$ .
So $P(A \cap B') = \frac{1}{3} - \frac{1}{12} = \frac{1}{4}$ .

Worth noting

$\frac{1}{2}$ is $P(A \cup B)$ , every outcome in either event.

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