Need help on a probability calculation Following is a problem statement:

A furniture shop has six identical steel cabinets of brand A and four
  identical steel cabinets of brand B. Three customers buy one cabinet
  each. Then the probability that two or more cabinets of brand A have
  been sold is

My answer was 1/2, but it's incorrect. My initial attempt was as follows 6/10*5/9 + 6/10*5/9*4/8 = 1/2.
 A: What is the probability that exactly two cabinets sold were of type A?
There are 3 possible situations:


*

*first customer buys A, second A, third buys B

*first customer buys B, second A, third buys A

*first customer buys A, second B, third buys A


The probability of 1. scenario is $\frac{6}{10}\cdot \frac{5}{9} \cdot \frac{4}{8}$
To determine the probability of the event that exactly two cabinets of type A have been bought, you have to add up the probability of all 3 scenarios, that is
$\frac{6}{10}\cdot \frac{5}{9} \cdot \frac{4}{8} + 
\frac{4}{10}\cdot \frac{6}{9} \cdot \frac{5}{8}+
\frac{6}{10}\cdot \frac{4}{9} \cdot \frac{5}{8} = \frac{1}{2}$
This is why your calculation failed.
If you add the probability that all cabinets were of type A, you get the probability you want.
$\frac{1}{2} +\frac{6}{10}\cdot \frac{5}{9} \cdot \frac{4}{8} = \frac{2}{3}$
A: Basic approach. (Assuming, as JMoravitz indicates in the comments, that each customer buys a cabinet from the available selection uniformly at random...)  There are ten cabinets: six A's and four B's.  Out of those ten, you choose three.  What expression represents the number of ways to choose three cabinets?  That's your denominator.
How many ways are there to choose three A's?  How many ways are there to choose two A's and one B?  Add those together and that's your numerator.
The probability is the numerator over the denominator.
