$P(A \cap B) = \frac14, P(\neg A) = \frac13, P(B) = \frac12, P(A \cup B) =$?

Question

My solution:

$P(A) = 1-P(\neg A) = \frac23$

$P(A \cap B) = P(A) \cdot P(B) = \frac23\cdot\frac12 = \frac26 = \frac13$

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac23+\frac12-\frac13 = \frac56$

But the correct answer is $\frac{11}{12}$. Why?

Answer 1

You correctly state that: $P(A\cup B) = P(A)+P(B)-P(A\cap B)$

Then you try to calculate $P(A\cap B)$ from $P(A)$ and $P(B)$ when (a) you do not know that they are independent and (b) you already have a value of $1/4$ given to you.

Which means that they are not independent, and you don't need to calculate it.

$$\mathsf P(A\cup B) = \tfrac 12+\tfrac 23-\tfrac 14 =\frac {11}{12}$$

Answer 2

Hints:

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

We also have that $\;P(\neg A)=1-P(A)\;$ , so...can you see why the answer sheet is correct?

As commented, if the events $\;A,\,B\;$ are independent then you can write $\;P(A\cap B)=P(A)P(B)\;$ , yet you're not given this.