# How to get the probability of a defective item

by Jeffrey   Last Updated May 15, 2019 18:20 PM

We are given three boxes as follows: Box $$1$$ has $$10$$ light bulbs of which $$4$$ are defective. Box $$2$$ has $$6$$ light bulbs of which $$1$$ is defective. Box $$3$$ has $$8$$ light bulbs of which $$3$$ are defective. If a box is selected at random and then a bulb is drawn. (a). What is the probability that the bulb is defective? (b). Draw a tree diagram to show the above question.

I have tried solving it. First, I thought that to select a box at random, it's gonna be $$3C1$$ which is three $$(3)$$. Then the total number of bulbs $$= 24$$ and there are $$8$$ defective bulbs. So the probability is $$3 \cdot\dfrac{8}{24} = 1$$. Please am I right?

Tags :

$$P(\text{defective}) = P(b_1)P(\text{defective}|b_1) + P(b_2)P(\text{defective}|b_2) + P(b_3)P(\text{defective}|b_3)$$
where $$b_1$$ means box 1 and so on. The idea is: you first have to chose a box $$b_1,b_2,b_3$$ at random, then you have to choose a bulb inside that box. Can you put together the numbers?