Question

A product code consists of four letters followed by five digits. Only the letters X, Y, and Z can be used ( and any of the digits 0,1,2,3,4,5,6,7,8,9 can be used). For example, XYXZ31513, ZYZZ90654, YYYY44444 are possible product codes.

If one such product code is chosen at random, what is the probability it includes the letter Y?

Answer 1

Answer:

$\frac{49}{81}$, 60.494%

Step-by-step explanation:

I am going to try to tackle this. I might miss a step as I am currently taking discrete mathematics.

We want to find all the possibilities with 1y, 2ys, 3ys, and 4ys, out of all total possibilities.

_ _ _ _   _ _ _ _ _

For the letters, in each spot, we have 3 choices. For the numbers, we have 10 choices in each spot

so

3*3*3*3 * 10*10*10*10*10

= 81*100000

= 8100000

The number of Y's will never affect the number of permutations of the numbers.

So:

For 1 y, we have

Y _ _ _ * 10*10*10*10*10

But we also have

_ Y _ _

_ _ Y _

and

_ _ _ Y

So we can multiply the number we get in one calculation by 4

4 *

Y _ _ _ * 10*10*10*10*10

2*2*2 * 10*10*10*10*10

= 800000 * 4  = 3200000

For 2 y's, we have

YY _ _

_ YY _

__ YY

3*

YY 2*2 * 10^5

4 * 100000

400000 * 3 = 1200000

For 3 y's, we have

YYY _

_ YYY

2 *

YYY 2 * 10^5

200000 * 2

= 400000

For 4 y's, we have

YYYY * 10^5

100000

Now we can add them all up and divide it by our original number 8100000

$\frac{100000 + 400000 + 1200000 + 3200000}{8100000} = \frac{49}{81}$