Thursday, January 17, 2013

Answer to the previous blog

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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.
  1. Given n,k,m,l, what is the probability mass function (PMF) for l? What is the expected value?
  2. Given n,m,l, what is the probability mass function (PMF) for k? What is the expected value?
(Sorry that I have to change the notations, so the presentation is clearer.)
Answer:
  1. 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 $$E[L|K=k]=\sum_{k=0}^k\frac{lC_l^kC_{m-l}^{n-k}}{C_m^n}.$$
  2. 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 $$P(\mbox{he finds all})=\begin{cases} \frac{C_{m-k}^{n-k}}{C_m^n} & n-k>m>k \\ \frac{C_{m-k}^{n-k}+C_{m-(n-k)}^k}{C_m^n}& m\ge n-k,k \\ \frac{C_{m-(n-k)}^k}{C_m^n} & k>m>n-k \end{cases}$$