Probability of choosing a defective appliance Amongst $15$ appliances there are $10$ functional and $5$ defective. We choose $3$ appliances at random. What is the probability of
A) all of them being functional 
B) exactly one being defective 
The total number of possible choices is $15\cdot 14\cdot 13$, number of choices containing only functional appliances is $10\cdot 9\cdot 8$ so 
$$P(A)=\frac{10\cdot 9\cdot 8}{15\cdot 14\cdot 13}$$
As for B), I tried a similar approach. We have $5$ possibilities of choosing a defective appliance and choose the other two from the functional ones so
$$P(B)=\frac{5\cdot 10\cdot 9}{15\cdot 14\cdot 13}$$
This is wrong however. I know that I am missing something but I am not sure why exactly this approach does not work. Thanks.

The number of ways to choose $1$ out of $5$ defective appliances is $\binom{5}{1}=5$
The number of ways to choose $2$ out of $10$ functional appliances is $\binom{10}{2}=45$
The number of ways to choose $3$ out of $15$ any type of appliances is $\binom{15}{3}=455$
Hence the probability of choosing $1$ defective and $2$ functional is $\frac{5\cdot45}{455}=\frac{45}{91}\approx0.5$
