The stock of a warehouse consists of boxes of high, medium and low quality light bulbs in respective proportions 1:2:2.

Question

The stock of a warehouse consists of boxes of high, medium and low quality light bulbs in respective proportions $1:2:2$. The probabilities of bulbs of three types being unsatisfactory are $0.3, 0.1$ and $0.2$ respectively. If a box is chosen at random and two bulbs in it are tested and found to be satisfactory, what is the probability that it contains bulbs of high quality.

I was thinking of solving through Bayes theorem. I calculated $P($high quality $|$ satisfactory$)$ from Bayes theorem but couldn't proceed further.

Answer 1

If you calculated $P(high|2satisf)$, then that is your answer. If you calculated $P(high|1satisf)$, then it's not quite right.

            / 2 satisfactory bulbs .3^2    0.018
/ high  1/5 -

            / 2 satisfactory bulbs .1^2   .004
- med   2/5 -

            / 2 satisfactory bulbs .2^2   .016
\ low   2/5 - 

Then $$\begin{split}P(high|2satis)&=\frac{P(high,2satisf)}{P(2satis)}\text { Bayes theorem}\\ &=\frac{P(high,2satisf)}{P(2satis,high)+P(2satis,med)+P(2satis,low)}\\ &=\frac{.018}{.018+.004+.016}\\ &=\frac 9 {19}\end{split}$$

So given that you have gotten 2 satisfactory bulbs out of the selected box, the probability that it is a high-quality box is $9/19$. This is what P(high | 2 satisfactory) means, e.g. given that 2 satisfactory bulbs are observed from the box, the probability that it was a high quality box is P(high | 2 satisfactory).