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Suppose there are

*n*lightbulbs,

*k*of them have defects. A technician is allowed to check

*m*lightbulbs and it turns out that exactly

*l*defected ones among those checked.

- Given
*n,k,m,l*, what is the probability mass function (PMF) for*l*? What is the expected value? - Given
*n,m,l*, what is the probability mass function (PMF) for*k*? What is the expected value?

Answer:

- Probability is (number of particular outcomes)/(number of all possible outcomes).
*k≥n*: In this case 0*≤l≤m*. The total number of ways for choosing*m*bulbs from the*n*bulbs is $C_m^n$. Number of ways that there are*l*defected bulbs among*m*chosen bulbs equals $C_l^kC_{m-l}^{n-K}$. Therefore the conditional probability for $L=l$ given $K=k$ is $$P(L=l|K=k)=\frac{C_l^kC_{m-l}^{n-k}}{C_m^n}.$$ Interestingly, rearranging the terms, we have (what does it mean?) $$P(L=l|K=k)=\frac{C_l^mC_{k-l}^{n-m}}{C_k^n}$$ The expected value equals the sum of (outcome × probability). Therefore $$E[L|K=k]=\sum_{k=0}^m\frac{lC_l^kC_{m-l}^{n-k}}{C_m^n}.$$*k≤n*: In this case 0*≤l≤k*. Again $$P(L=l|K=k)=\frac{C_l^kC_{m-l}^{n-k}}{C_m^n}.$$ The expected value is $$E[L|K=k]=\sum_{k=0}^k\frac{lC_l^kC_{m-l}^{n-k}}{C_m^n}.$$

- Indeed there is something missing in the question: How is
*k*distributed? Lets assume that it is*K~Bin(N,p)*, that is, each bulb has a probability*p*to be defected and that $P(K=k)=C_k^np^k(1-p)^{n-k}$. Therefore $$P(K=k|L=l)=\frac{P(Y=K, L=l)}{P(L=l)} =\frac{P(K=k)P(L=l|K=l)}{\sum_{k=l}^nP(K=k)P(L=l|K=k)}$$ and $$E[K|L=1]=\frac{kP(K=k)P(L=l|K=l)}{\sum_{k=l}^nP(K=k)P(L=l|K=k)}.$$

The second question:

Suppose there are

*n*lightbulbs,

*k*of them have defects. A technician is allowed to check

*m*lightbulbs. What is the probability that he can find out all the defected?

Answer: The technician can find all the defected bulbs if either all the defected bulbs or all the normal bulbs are among those checked. Therefore the answer is $$P(\mbox{he finds all})=\begin{cases} \frac{C_{m-k}^{n-k}}{C_m^n} & n-k>m>k \\ \frac{C_{m-k}^{n-k}+C_{m-(n-k)}^k}{C_m^n}& m\ge n-k,k \\ \frac{C_{m-(n-k)}^k}{C_m^n} & k>m>n-k \end{cases}$$