Suppose that 10% of all bits transmitted through a digital communication channel are erroneously received and that whether any particular bit is erroneously received is independent of whether any other bit is erroneously received. Consider sending a very large number of messages, each consisting of 20 bits.

a. What proportion of these messages will have at most 2 erroneously received bits?

b. What proportion of these messages will have at least 5 erroneously received bits?

c. For what proportion of these messages will more than half the bits be erroneously received?

Answers

Answer 1

Answer:

a) 0.677

b) 0.0431

c) 0

Step-by-step explanation:

For each bit, there are only two possible outcomes. Either it is correctly received, or it is erroneously received. Whether any particular bit is erroneously received is independent of whether any other bit is erroneously received. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

10% of all bits transmitted through a digital communication channel are erroneously

This means that [tex]p = 0.1[/tex]

Messages of 20 bits:

This means that [tex]n = 20[/tex]

a. What proportion of these messages will have at most 2 erroneously received bits?

This is [tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{20,0}.(0.1)^{0}.(0.9)^{20} = 0.1216[/tex]

[tex]P(X = 1) = C_{20,1}.(0.1)^{1}.(0.9)^{19} = 0.2702[/tex]

[tex]P(X = 2) = C_{20,2}.(0.1)^{2}.(0.9)^{18} = 0.2852[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1216 + 0.2702 + 0.2852 = 0.677[/tex]

b. What proportion of these messages will have at least 5 erroneously received bits?

This is

[tex]P(X \geq 5) = 1 - P(X < 5)[/tex]

In which

[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{20,0}.(0.1)^{0}.(0.9)^{20} = 0.1216[/tex]

[tex]P(X = 1) = C_{20,1}.(0.1)^{1}.(0.9)^{19} = 0.2702[/tex]

[tex]P(X = 2) = C_{20,2}.(0.1)^{2}.(0.9)^{18} = 0.2852[/tex]

[tex]P(X = 3) = C_{20,3}.(0.1)^{3}.(0.9)^{17} = 0.1901[/tex]

[tex]P(X = 4) = C_{20,4}.(0.1)^{4}.(0.9)^{16} = 0.0898[/tex]

[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.1216 + 0.2702 + 0.2852 + 0.1901 + 0.0898 = 0.9569[/tex]

Then

[tex]P(X \geq 5) = 1 - P(X < 5) = 1 - 0.9569 = 0.0431[/tex]

c. For what proportion of these messages will more than half the bits be erroneously received?

This is [tex]P(X \geq 11)[/tex]

[tex]P(X = 11) = C_{20,11}.(0.1)^{11}.(0.9)^{9} \approx 0[/tex]

If 11 is approximately 0, the rest(11, 12, 13,...) will also be approximately 0. So this proportion is 0.


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Answers

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Here we will show you step-by-step with detailed explanation how to calculate 452 divided by 6 using long division.

Before you continue, note that in the problem 452 divided by 6, the numbers are defined as follows:

452 = dividend

6 = divisor

Step 1:

Start by setting it up with the divisor 6 on the left side and the dividend 452 on the right side like this:

6 ⟌ 4 5 2

Step 2:

The divisor (6) goes into the first digit of the dividend (4), 0 time(s). Therefore, put 0 on top:

0

6 ⟌ 4 5 2

Step 3:

Multiply the divisor by the result in the previous step (6 x 0 = 0) and write that answer below the dividend.

0

6 ⟌ 4 5 2

0

Step 4:

Subtract the result in the previous step from the first digit of the dividend (4 - 0 = 4) and write the answer below.

0

6 ⟌ 4 5 2

- 0

4

Step 5:

Move down the 2nd digit of the dividend (5) like this:

0

6 ⟌ 4 5 2

- 0

4 5

Step 6:

The divisor (6) goes into the bottom number (45), 7 time(s). Therefore, put 7 on top:

0 7

6 ⟌ 4 5 2

- 0

4 5

Step 7:

Multiply the divisor by the result in the previous step (6 x 7 = 42) and write that answer at the bottom:

0 7

6 ⟌ 4 5 2

- 0

4 5

4 2

Step 8:

Subtract the result in the previous step from the number written above it. (45 - 42 = 3) and write the answer at the bottom.

0 7

6 ⟌ 4 5 2

- 0

4 5

- 4 2

3

Step 9:

Move down the last digit of the dividend (2) like this:

0 7

6 ⟌ 4 5 2

- 0

4 5

- 4 2

3 2

Step 10:

The divisor (6) goes into the bottom number (32), 5 time(s). Therefore put 5 on top:

0 7 5

6 ⟌ 4 5 2

- 0

4 5

- 4 2

3 2

Step 11:

Multiply the divisor by the result in the previous step (6 x 5 = 30) and write the answer at the bottom:

0 7 5

6 ⟌ 4 5 2

- 0

4 5

- 4 2

3 2

3 0

Step 12:

Subtract the result in the previous step from the number written above it. (32 - 30 = 2) and write the answer at the bottom.

0 7 5

6 ⟌ 4 5 2

- 0

4 5

- 4 2

3 2

- 3 0

2

You are done, because there are no more digits to move down from the dividend.

The answer is the top number and the remainder is the bottom number.

Therefore, the answer to 452 divided by 6 calculated using Long Division is:

75

2 Remainder

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