Thread: Probability question involving bank PIN numbers. P

"Bank of america customers select their own personal three digit identification for use at ATM's. What is the probability that Mr. Jones and Mrs. Smith select the same PIN?"

My approach is >

Since we have the PIN is made of three digits, for each digit, we could choose any number from 0-9. So the number of all possible PIN's that could be selected is 10x10x10=1000
The restriction is that Mr. Jones and Mrs. Smith will have the PIN. They could have any PIN, but it must be the same. So, one of them has the choice to choose "any" PIN. The other one has to choose the same PIN.

Hence if P(Smith)= Probability of Smith choosing any PIN=1
and if P(Jones)= Probability of Jones choosing PIN chosen by smith=(1/1000)

Then, P(Smith and Jones)= Probability of smith and Jones choosing the same pin=(1)(1/1000) =1/1000

Is this correct???

Look good to me, since the number chosed by each other seems independant event P(A and B) = P(A)*P(B)