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?

Someone please help including the tree diagram

$P(\text{defective}) = P(b_1)P(\text{defective}|b_1) + P(b_2)P(\text{defective}|b_2) + P(b_3)P(\text{defective}|b_3)$

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?

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