Bunuel wrote:

Official Solution:

A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digits 6?

A. \(\frac{860}{90,000}\)

B. \(\frac{810}{100,000}\)

C. \(\frac{858}{100,000}\)

D. \(\frac{860}{100,000}\)

E. \(\frac{1530}{100,000}\)

Total # of 5-digit codes is \(10^5\), notice that it's not \(9*10^4\), since in a code we can have zero as the first digit.

# of passwords with three digits 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have \(9*9\) and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.

\(P=\frac{favorable}{total}=\frac{810}{10^5}\).

Answer: B

Hi

Bunuel,

I did exactly the same except I could not do the last part i.e arrangement of 3-6's among 5 places. Am I correct in saying that we have to find the total no of arrangements for A666A & A666B? Where A and B are different digits. So in this case {5!/(3!2!) + 5!/3!} depending on whether remaining two digits are same or different.