Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 18

All size-six subsets are equally likely under uniform sampling without replacement. There are $\left(\genfrac{}{}{0ex}{}{12}{6}\right)$ subsets in total, and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)\left(\genfrac{}{}{0ex}{}{6}{3}\right)$ contain the required three members from each group. Thus the hypergeometric distribution gives

Let $A,B,C$ denote the mutually exclusive supplier events and $D$ the defective-bulb event. Their probabilities and conditional defect rates areThese supplier events partition the possible bulbs. The following calculations use the law of total probability and Bayes' theorem with that partition.

The law of total probability givesFor $\alpha >0$, Bayes' theorem therefore yieldsAt the allowed endpoint $\alpha =0$, no bulb is defective: $\mathbb{P}(D)=0$. This conditional probability is then undefined, rather than $2/5$. The value $2/5$ is its limit as $\alpha \to 0^{+}$, but does not define conditioning on the impossible event within the stated discrete model.

The non-defective event is the complement $D^c$. Using the total defective probability found above,This formula is valid throughout $0\le \alpha \le 1$, and its value is at least $1/2$.

The joint probability of supplier $C$ and a non-defective bulb is $\mathbb{P}(C\cap D^c)=(2/5)(1-\alpha /4)$. Normalize by the positive non-defective probability:This is defined even at $\alpha =0$, where it reduces to the original supplier fraction $2/5$. At $\alpha =1$ it is $3/5$.