" A red box contains $8$ items, of which $3$ are defective, and a blue box contains $5$ items, of which $2$ are defective. An item is drawn at random from each box. What is the probability that one item is defective and one is not?"
I interpret the problem can be solved by multiplying the probability that an item from the red box is defective by the probability that an item from the red box is not defective (with the knowledge that I do not return items back into the red box after picking the first item) plus the probability that an item from the blue box is defective times the probability that an item from the blue box is not defective (with the same knowledge as before that I do not return the items back into the blue box after picking the first item.
I.E. I believe that one solves the probability with $\frac 38\cdot \frac 57 + \frac 25 \cdot \frac 34$.
How is my interpretation of the problem incorrect?
From Barron's SAT Subject Test Math 2 book
Best Answer
You almost have the correct approach.
Let $R$ be the event of drawing a defective from the red box. $\mathsf P[R]=\dfrac{3}{8}$
Let $B$ be the event of drawing a defective from the blue box. $\mathsf P[B]=\dfrac 2 5$
You draw one item independently from each separate box.
Find: $\quad\mathsf P[(R\cap \neg B) \cup (\neg R \cap B)] \\[2ex] = \mathsf P[R]\;\mathsf P[\neg B]+\mathsf P[\neg R]\;\mathsf P[B] \\[2ex] = \dfrac{3}{8}\cdot\dfrac{3}{5}+\dfrac{5}{8}\cdot\dfrac{2}{5} $
This is $\mathsf P[R_1]\,\mathsf P[\neg R_2\mid R_1]+\mathsf P[B_1]\,\mathsf P[\neg B_2\mid B_1]$, Where $R_n$ is the event that the $n$-th item drawn from the red box is defective.
Why would you do that? You need to draw one item from each box.
Let alone, why are you adding? Those are not exclusive events.