two ways to solve this kind of problem
way 1. count the number of event needed and number of event available
number of events available is 6*5*4/3*2*1 (set without order-Princeton review
). Number is 20
number of events needed is: we take Tom and 2 persons from 4 person. there are 4*3/2 (set without order). number is 6
result is 6/20
way 2: separate into cases, find P of each case and multiple or plus together.
case 1: find P (A and first person and second person)
case 2, find P (first person and A and second person
case 3, find P (first person and second person and A)
As you noticed that all 3 probabilities are the same and they are the same for a reason.
Consider this example what is the probablity of selecting the integer 1 out of 4 different numbers, it's 1/4 and what is the probability of selecting 2 out of 4 different numbers, it's also 1/4.
The same principle is applicable to this problem. The order in which you select a person, first, second or third does not matter, the probability is the same.
Hard work is the main determinant of success