Binomial Probability Problem

Question

This is problem 8.11 in the Marcel Finan book.

A store has 80 modems in its inventory, 30 coming from Source A and the remainder from Source B. Of the modems from Source A, 20% are defective. Of the modems from Source B, 8% are defective. Calculate the probability that exactly two out of a random sample of five modems from the store’s inventory are defective.

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$$P(\text{defective}) = \frac{3}{8} \frac{1}{5} + \frac{5}{8} \frac{8}{100}=\frac{1}{8}$$

$$P(\text{exactly 2 defective}) = \binom{5}{2} P(\text{defective})^2 (1-P(\text{defective}))^3 = 0.104675$$

But the answer key says 0.102. What have I done wrong?

Answer 1

Consider all modems in the store totalling $80$.

Once they are comingled, it doesn't matter where they come from.

$$\begin{array}\\ \hline\ &\text{Source} &A &B &&\text{Total}\\ \hline &\text{Total} &30 &50 &&80\\ &\% \text { Defective} &20\% &8\%\\ &\text{Defective} &\;\ 6 &\;\ 4 &&10\\ &\text{Not Defective} &24 &46 &&70\\ \hline \end{array}$$

Total number defective = $10$.

Total number non-defective = $70$.

Choose $5$ modems.

Probability of getting exactly $2$ defective is:

$$\frac{\binom {10}2\binom {70}3}{\binom {80}5}=0.102\qquad\blacksquare$$

Answer 2

You're treating the defects as independent, but they're not, since if you find a defect in a modem, that modem is somewhat more likely to be from source $A$, so the remaining modems are somewhat more likely to be from source $B$.

Answer 3

A store has 80 modems in its inventory, 30 coming from Source A and the remainder from Source B. Of the modems from Source A, 20% are defective. Of the modems from Source B, 8% are defective. Calculate the probability that exactly two out of a random sample of five modems from the store’s inventory are defective.

The problem is that it is not exactly a "Binomial Probability Problem", because the count of defectives in the sample of five is not independent of how many of the sample are sourced from each factory.

$\mathsf P(D=2) = \dfrac{\binom{30}{0}\binom{50}{5}\binom{5}{2}0.08^20.92^3+\binom{30}{1}\binom{50}{4}\Big(\binom{1}{1}0.2\binom{4}{1}0.08\cdot0.92^3+\binom{1}{0}0.8\binom{4}{2}0.08^20.92^2\Big)+\binom{30}{2}\binom{50}{3}\Big(\binom{2}{2}0.2^2\binom{3}{0}0.98^3+\binom{2}{1}0.2\cdot0.8\binom{3}{1}0.08\cdot0.98^2+\binom{2}{0}0.8^2\binom{3}{2}0.08^20.98\Big)+\binom{30}{3}\binom{50}{2}\Big(\binom{3}{2}0.2^20.8\binom{2}{0}0.98^2+\binom{3}{1}0.2\cdot0.8^2\binom{2}{1}0.08\cdot0.98+\binom{3}{0}0.8^3\binom{2}{2}0.08^2\Big)+\binom{30}{4}\binom{50}{1}\Big(\binom 4 2 0.2^20.8^2\binom 1 0 0.98+\binom{4}{1}0.2\cdot0.8^3\binom 1 0 0.08\Big)+\binom{30}5\binom{50}0\binom 5 20.2^20.8^3}{\binom{80}{5}} $

Which we approximate by a hypergeometric distribution

$$\mathsf P(D=2)~=~\dfrac{\dbinom{30\cdot 0.2+50\cdot0.08}{2}\dbinom{30\cdot 0.8+50\cdot0.92}{3}}{\dbinom{80}{5}}$$