<s>[INST] Find the probability that atleast one valve is defective. A factory A produces $10$% defective valves and another factory $B$ produces 20% defective valves.A bag contains $4$ valves of factory $A$ and $5$ valves of factory B.If two valves are drawn at random from the bag,find the probability that at least one valve is defective. 

$P(\text{at least one valve is defective})=\\=1-P(\text{none of the two valves are defective})=\\=1-\left(\frac{\binom{4}{2}}{\binom{9}{2}}(0.9)^2+\frac{\binom{5}{2}}{\binom{9}{2}}(0.8)^2+\frac{\binom{4}{1}\binom{5}{1}}{\binom{9}{2}}(0.9)(0.8)\right)=\frac{517}{1800}$,
but the answer given is $\frac{303}{1800}$ I don't know where i am wrong.
 [/INST] The answer is indeed $\frac{517}{1800}$, because:
$$P(\text{at least one of two defective})=1-P(\text{both normal})=\\
1-[P(ANAN)+P(ANBN)+P(BNAN)+P(BNBN)]=\\
1-\left[\frac{4}{9}\cdot \frac9{10}\cdot \frac38\cdot \frac9{10}+
\frac{4}{9}\cdot \frac9{10}\cdot \frac58\cdot \frac8{10}+
\frac{5}{9}\cdot \frac8{10}\cdot \frac48\cdot \frac9{10}+
\frac{5}{9}\cdot \frac8{10}\cdot \frac48\cdot \frac8{10}\right]=\\
1-\frac{1283}{1800}=\frac{517}{1800}.$$
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