<s>[INST] 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.
 [/INST] 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.
$\frac{1}{2} +\frac{6}{10}\cdot \frac{5}{9} \cdot \frac{4}{8} = \frac{2}{3}$
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