Perimeter of a sentence

Probability Level 2

In an imaginary language , there are only three distinct letters. For simplicity, we call them: " A ", " B " & " C ".

The following table illustrates the average frequency of characters across all sentences written in that language. In addition, a " space " is included in the table and all of these characters also have a distinct perimeter:

character	frequency	perimeter
"A"	0.36	2
"B"	0.21	3
"C"	0.29	1
"space"	0.14	0

An "average sentence" takes the above frequencies into account to apply them to the character count of such a sentence.

However, it should represent "real" or possible sentence. This means that the count of each character can only be a natural number . The character with the higher overall frequency would take the spot in a "tie".
If there were $221.2$ A's and $15.8$ B's in a sentence from a language with only "A" and "B" according to the frequency table of that language, the "average sentence" would contain $16$ B's and $221$ A's.

How big is the sum of all letters' perimeters in an "average sentence" of length 42?

43 2772 69 36 37 2940 70 41

1 solution

Ussar Niaem
Dec 31, 2018

Frequencies

$A = 42 * 0.36 = 15.12$

$B = 42 * 0.21 = 8.82$

$C = 42 * 0.29 = 12.8$

$spaces = 42 * 0.14 = 5.88$

Flooring these values, we already have $40$ "whole" characters, which means that we have to find the two characters nearest to their next integer respectively and add $1$ to those:

$A = 15$

$B = 9$

$C = 12$

$spaces = 6$

Now, we can just sum up the A's, B's and C's multiplied by their respective perimeters:

$15 \cdot 2 + 9 \cdot 3 + 12 = 69$

Perimeter of a sentence

How big is the sum of all letters' perimeters in an "average sentence" of length 42?

1 solution

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