thangvietnam wrote:

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)

P1=(1/6)*(4/5)*(3/4)=1/10

case 2, find P (first person and A and second person

P2=(4/6)*(1/5)*(3/4)=1/10

case 3, find P (first person and second person and A)

P3=(4/6)*(3/5)*(1/4)=1/10

P=p1+p2+p3=3/10

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.

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