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**HallsofIvy** There are 5 people, any one of which is equally likely to be on the committee. The probability any one of them, and in particular, Burke, is on it is 1/5.

(There are 5 choices for the first person, 4 for the second, 3 for the for the last so there are 5(4)(3)= 60 possible three person ways to form a three person committee from 5 people. If we require that Burke be on the committee, there at 4 choices for the next person and 3 for the last so there are 4(3)= 12 ways to choose a three person committee that includes Burke. The probability that the committee includes Burke is 12/60= 1/5.)