Practive Problem on Independence
Q: There are five volunteers(Called A,B,C,D,E), and there are four volunteer positions(Position 1, postion 2, position 3, position 4). A volunteer is equally likely to be place on one of these for positions.
1.What is the probability that five volunteers are in the same position? 2.Given that C and D is not in position 1, what is the probability that both A and B are in the position 1?
A:
1.
P1 = p(all in position 1) + p(all in position 2) + p(all in position 3) + p(all in position 4)
= [(1/4)^4]*(1/4) + [(1/4)^4]*(1/4) + [(1/4)^4]*(1/4) + [(1/4)^4]*(1/4)
= (1/4)^4
= 1/64
2.
p2 = p(A is in position 1|C and D is not in position 1)*p(B is in position 2|C and D is not in position 1) Event A is in position 1 is independent with the Event C and D is in position 1.
= p(A is in position 1) *p(C and D is not in position 1)(* p(B is in position 1)
= (1/4)*(1/4)*(1-(1/16)) = (1/16)*(15/16) = 15/256
That's Ruofei Chen's problem. What he did is correct except a small calculation error => (1/4)^4 = 1/256, not 1/64
