A computer wholesaler sells nine different computers and each is priced differently. If the wholesaler chooses three computers for display at a trade show, what is the probability (all things being equal) that the two most expensive computers will be among the three chosen for display?
Answers:
A) 15/56  B) 3/28  C) 1/12  D) 1/56  E) 1/168
Since, two of the choices are prefixed, we are free to choose 1 from the rest of the 7 avilable. so 7C1 is the numerator.   Total no of ways we can choose 3 from 9 is 9c3, which is the denominator.   So, the probability: 7c1/9c3= 1/12 ans is 1/12.
<result>C</result>
A computer wholesaler sells nine different computers and each is priced differently. If the wholesaler chooses three computers for display at a trade show, what is the probability (all things being equal) that the two most expensive computers will be among the three chosen for display?