Question

Let us assume that we are allowed to print so called College Money used for shopping only within campus. The money is composed a unique serial numbers like M20130BC. It always starts with M, follows with five digital numbers chosen from 0 to 9, and then two letters chosen from A to Z. a. What is the total number of such series numbers? b. What is the probability of getting the serial number M88888MM? C. What is the probability of getting a serial number whose last two letters to be AA?

Answer 1

Answer:

a. 67600000

b. 0.0001396

c. 0.001479

Step-by-step explanation:

a. Since it is a permutation with repetition the formula is:

$n^r$

Where $n$ is the number of optionss, here you will use $n_1=10$ and $n_2=26$, one for the numbers and the other for the letters. The exponent $r$ corresponds to how many times you have to choose a number (5 times) or a letter (two times because the first letter is always the same), then:

$n_1^5*n_2^2=10^5*26^2=100000*676=67600000=6.76\times10^7$

b. The probability of getting that exactly serial number is 1 divided by the total of possibilities:

$P_b=\frac{1}{6.76\times10^7}=0.0001396=1.396\times10^{-4}$

C. The probability is the number of serials ending with AA divided by the total. The serial ending with AA can have any numbers then  $n=10$ and $r=5$

$n^r=10^5=100000=1\times10^5$

$P_c=\frac{1\times10^5}{6.76\times10^7}=0.001479=1.479\times10^{-3}$