Question

HELP!!!! Злой экзаменатор никогда не ставит пятерок по информатике. По причине своей зловредности он заранее определил количество отметок каждого вида и произвольно расставил их абитуриентам. Количество информации, содержащееся в сообщении «Абитуриент Иванов не провалился на экзамене», равно log23 бит. Информационный объем сообщения «Абитуриент Сидоров получил тройку» равен двум битам. 22 абитуриента получили двойку или тройку. Найти количество абитуриентов, сдавших информатику.

Answer 1

Формула Шеннона: I=log2 N  (I-кол-во инф-ции; N=всё/нужное)
---------------------------------------------------------------------------------------
обозначение:
ч - кол-во четвёрок
т - кол-во троек
д - количество двоек
пятёрок нет
-----------------------------------------------------------------------------------------
по условию   т+д=22
не провалившиеся = ч+22  ----> log2 3=log2 ((ч+22)/(ч+т))

инф-ный объём сообщения, что тройка = 2 битам  ---->
   2=log2 ((ч+22)/т)
---------------------------------------------------------------------------------
Имеем систему трёх уравнений с тремя неизвестными:
т+д=22
log2 3=log2 ((ч+22)/(ч+т))
2=log2 ((ч+22)/т)                                ---->log2 4=log2 ((ч+22)/т)
       Преобразуем:   
..................................система:
т+д=22
3=(ч+22)/(ч+т)
4=(ч+22)/т        ------>   ч+22=4т
.................................
т+д=22
3ч+3т=ч+22
4т=ч+22
...........................
т+д=22
3ч+3т=4т
4т=ч+22
.........................
т+д=22
3ч=т
12ч=ч+22
....................
т+д=22
3ч=т
ч=2
.............
ч=2                   т=3ч=6                    д=22-6=16
всего:   2+6+16=24  абитуриента
         ответ  24

Answer 2

Problem Analysis

We are given that a malicious examiner has predetermined the number of marks for each type and randomly assigned them to the students. The information contained in the message "Student Ivanov did not fail the exam" is equal to log2(3) bits, and the information volume of the message "Student Sidorov received a grade of three" is equal to 2 bits. We are also told that 22 students received a grade of two or three. We need to find the number of students who passed the computer science exam.

Solution

Let's assume that the number of students who passed the computer science exam is x.

From the given information, we can deduce the following equations:

1. The information contained in the message "Student Ivanov did not fail the exam" is equal to log2(3) bits. 2. The information volume of the message "Student Sidorov received a grade of three" is equal to 2 bits. 3. 22 students received a grade of two or three.

Using these equations, we can set up the following system of equations:

Equation 1: log2(3) = log2(x) - log2(22 - x) Equation 2: 2 = log2(x) - log2(22 - x)

To solve this system of equations, we can eliminate the logarithms by raising 2 to the power of both sides:

Equation 1 (after exponentiation): 3 = x / (22 - x) Equation 2 (after exponentiation): 4 = x / (22 - x)

Now, we can solve these equations simultaneously:

Equation 1: 3(22 - x) = x Equation 2: 4(22 - x) = x

Simplifying these equations:

Equation 1: 66 - 3x = x Equation 2: 88 - 4x = x

Solving for x:

Equation 1: 66 = 4x x = 66 / 4 x = 16.5

Equation 2: 88 = 5x x = 88 / 5 x = 17.6

Since the number of students cannot be fractional, we round down the value of x to the nearest whole number:

x = 16

Therefore, the number of students who passed the computer science exam is 16.

Answer

The number of students who passed the computer science exam is 16.