Question

a product code consists of picking a two-digit number (including 00) and 3 not necessarily distinct letters. How many codes are possible? how many codes contain at least one 9? What is the probability that a code does not contain the number 9?

Answer 1

Answer:

The codes are:

N,_,_,_

Where each _ represents a possible letter, out of 26.

And N is a number of two digits.

So in total we can think that we have 5 empty slots.

_,_,_,_,_

In the first slot we can put any digit between 0 and 9, so here we have 10 options.

In the second slot we can put any digit between 0 and 9, so here we have 10 options.

In the third slot we can put any letter, so here we have 26 options (and exactly the same for the fourth and fifth slots)

The total number of different combinations is equal to the product of the number of options for each slot.

C = 10*10*26*26*26 = 1,757,600

Now, if we want to have at least one nine, we can fix it in the first slot.

Then we have:

in the first slot we have only one option (the 9)

In the second slot we can put any digit between 0 and 9, so here we have 10 options.

In the third slot we can put any letter, so here we have 26 options (and exactly the same for the fourth and fifth slots)

The number of combinations is:

C = 1*10*26*26*26 = 175,760

But we also should consider the case where we fix the 9 in the second slot, so the actual number of combinations is twice the number above.

C = 2*175,760 = 351,520

The probability that the code does not contain the number 9.

Now in the first slot we have only 9 options, all the whole numbers between 0 and 8.

The same for the second slot, 9 options.

For the third, fourth and fifth slot is the same as before.

The total number of combinations is:

C = 9*9*26*26*26 = 1,423,656