A computer wholesaler sells 11 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? Pick
A) 15/56  B) 3/28  C) 1/28  D) 1/56  E) 3/55
Since, two of the choices are prefixed, we are free to choose 1 from the rest of the 9 avilable. so 9C1 is the numerator.   Total no of ways we can choose 3 from 11 is 11c3, which is the denominator.   So, the probability: 9c1/11c3= 3/28 ans is 3/55.
<result>E</result>
A computer wholesaler sells 11 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?