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.
 A: The Probability that factory $A$ produces defective values is $\dfrac{10}{100}=\dfrac{1}{10}$
The Probability that factory $B$ produces defective values is $\dfrac{20}{100}=\dfrac{1}{5}$
Given a bag contains $4$ values of factory $A$ and $5$ values of factory $B$ and two values are drawn random.
$$P(\mbox{at least one defective})=1-P(\mbox{both are non-defective})$$
$$P(\mbox{both are non-defective})=P(\mbox{both values of factory }B)\times P(\mbox{both are non-defective})+P(\mbox{both values of factory }B)\times P(\mbox{ both are non defective})+P(\mbox{one value of $A$ and other of factory $B$})\times P(\mbox{both are nondefective})$$
$$=\dfrac{\dbinom{4}{2}}{\dbinom{9}{2}}\left(\frac{9}{10}\right)^2+\dfrac{\dbinom{5}{2}}{\dbinom{9}{2}}\left(\dfrac45\right)^2+\dfrac{\dbinom{4}{1}\cdot\binom{5}{1}}{\binom{9}{2}}\times\left(\frac{9}{10}\right)\times\left(\frac{4}{5}\right)$$
$$=\dfrac{27}{200}+\dfrac{8}{45}+\dfrac{2}{5}=\dfrac{1283}{1800}$$
Now, $P(\mbox{at least one defective})=1-\dfrac{1283}{1800}=\approx0.29$
So, the answer what you got is $\dfrac{517}{1800}\approx0.29$ which is correct.
A: The answer is indeed $\frac{517}{1800}$, because:
$$P(\text{at least one of two defective})=1-P(\text{both normal})=\\
1-[P(ANAN)+P(ANBN)+P(BNAN)+P(BNBN)]=\\
1-\left[\frac{4}{9}\cdot \frac9{10}\cdot \frac38\cdot \frac9{10}+
\frac{4}{9}\cdot \frac9{10}\cdot \frac58\cdot \frac8{10}+
\frac{5}{9}\cdot \frac8{10}\cdot \frac48\cdot \frac9{10}+
\frac{5}{9}\cdot \frac8{10}\cdot \frac48\cdot \frac8{10}\right]=\\
1-\frac{1283}{1800}=\frac{517}{1800}.$$
