Question

A bankcard "password" is a sequence of 4 digits. (a) how many passwords are there? (b) how many have no repeated digit? (c) how many have the digit "5" repeated j times for all possible values of j? (d) if a password were generated at random, what would be the expected number of 5s?

Answer 1

There are 10 digits. These are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The password needs to be 4 digits long.
(A) How many passwords are there?
Since there are no restrictions on the digits we can use, there are 10 choices for the first, 10 for the second, 10 for the third and 10 for the fourth. Whatever digit we use for a particular space it does not affect the other. We say they are independent. Therefore, we can find the number of passwords by using the counting principle. The counting principle states that for independent events, if there are m ways that an event A can happen and n ways that an event B can happen, the number of ways they both can happen is mn (multiply them). Therefore, the answer here is 10 x 10 x 10 x 10 = 10000. There are 10,000 passwords.

(B) For an explanation of the counting principle see part (A) which is used to solve these problems see part A. Now, here the digits cannot repeat so there are 10 choices for the first but only 9 for the second (since we can't pick the number we just picked again). That leaves 8 choices for the next number and 7 for the last. We multiply these to get: 10 x 9 x 8 x 7 = 5040. There are 5,040 passwords where the digits do not repeat.
(c) Here we are asked for all the passwords that contain a repetition of the number 5. The number five can be repeated 2 times, 3 times or 4 times since there are only 4 digits used in the password.
Repeated two times --> These are passwords that look like this:55XX5XX5XX555X5XX5X5X55X
There are six  ways to do it but all of them occur the same number of times because they are all two fives and two non-fives. So taking 55XX we have 1 choice for the first number (it has to be a five), 1 choice for the second, 9 choices for the next (anything but 5), and 9 for the last. That makes (1)(1)(9)(9) = 81 ways. As there are 6 ways to get two fives that makes (6)(81)=486 ways
Repeated three times --> passwords look like this:555X55X55X55X55There are four ways to get three fives. Each occurs this many times...let's use 555X so there is 1 choice for the first number (it has to be a five), 1 choice for the next, 1 choice for the next and 9 choices for the last. That's (1)(1)(1)(9) = 9 ways. Since we can get three fives in 6 different ways that's (9)(6)=54 passwords.
Repeated four times --> there's only one password and it is 5555. This can only happen one way (everything is a 5).
So we take the number of passwords with two repeated 5s (486) and add those with three repeated 5s (54) and then the one with all fives (1) to get our answer.
There are 486+54+1= 541 passwords where the number 5 repeats either twice, three times or four times.
(D) Let's find out how many plates have no fives. This would be (9)(9)(9)(9)=6561.
The number of plates with one five is (1)(9)(9)(9)=729

So now let's find the expected value for the number of fives that appear. We do this by finding P(no fives)(0)+P(one five)(1)+P(two fives)(2)+P(3 fives)(3)+P(4 fives)(4)

There are 10,000 passwords all together. So to find the probabilities we do this: (# of passwords with however many fives)/10000
That gives: (6561/10000)(0)+(729/10000)(1)+(486/10000)(2)+(54/1000)(3)+(1/10000)(4)= (729/10000)+(972/10000)+(162/10000)+(4/10000)=1876/10000=.1867 Therefore, it is expected that if a random plate is selected it has .1876 fives. That means the overwhelming majority of the time there isn't even one five.