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Benjamincompton07
27.11.2020 •
Mathematics
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, 44444 are possible product codes. If one such product code is chosen at random, what is the probability it includes the letter Y?
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Ответ:
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
Ответ: