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Law Incorporation [45]
4 years ago
12

Two spies have to communicate using secret code. They need to create exactly 30 possible precoded messages, using a single numbe

r and letter. Which structure should the code have? A. Select a number from {1, 2, 3, 4} and a vowel
B. Select a number from {1, 2, 3, 4, 5} and a vowel.

C. Select a number from {1, 2, 3, 4, 5, 6} and a vowel.

D. Select a number from {1, 2, 3, 4, 5} and a consonant
Mathematics
2 answers:
amid [387]4 years ago
3 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

<span>The structure should the code have should be letter C which is </span><span>Select a number from {1, 2, 3, 4, 5, 6} and a vowel.
</span>
denis-greek [22]4 years ago
3 0

Answer:

The code should have the structure C.

Step-by-step explanation:

Given : Two spies have to communicate using secret code. They need to create exactly 30 possible precoded messages, using a single number and letter.

To Find: Which structure should the code have?

Solution:

The structure that have 30 possible outcomes , the code should have that structure .

Option A:  Select a number from {1, 2, 3, 4} and a vowel

Since we are given that the code contains one number and one letter

No.of vowels = {a,e,i,o,u}=5

Out of these five we will choose only one

We are supposed to choose a number from {1, 2, 3, 4}

Out of these four numbers we will choose only one

Now to find no. of possible outcomes we will use combination

Formula : ^nC_r=\frac{n!}{r!(n-r)!}

So, no. of possible outcomes from option A  :

^5C_1\times ^4C_1

\frac{5!}{1!(5-1)!} \times \frac{4!}{1!(4-1)!}

\frac{5!}{1!(4)!} \times \frac{4!}{1!(3)!}

5 \times 4

20

Thus no. of possible outcomes from Option A is 20.

Option B: Select a number from {1, 2, 3, 4, 5} and a vowel.

Since we are given that the code contains one number and one letter

No.of vowels = {a,e,i,o,u}=5

Out of these five we will choose only one

We are supposed to choose a number from {1, 2, 3, 4,5}

Out of these five numbers we will choose only one

So, no. of possible outcomes from option B  :

^5C_1\times ^5C_1

\frac{5!}{1!(5-1)!} \times \frac{5!}{1!(5-1)!}

\frac{5!}{1!(4)!} \times \frac{5!}{1!(4)!}

5 \times 5

25

Thus no. of possible outcomes from Option B is 25.

Option C: Select a number from {1, 2, 3, 4, 5, 6} and a vowel.

Since we are given that the code contains one number and one letter

No.of vowels = {a,e,i,o,u}=5

Out of these five we will choose only one

We are supposed to choose a number from {1, 2, 3, 4,5,6}

Out of these six numbers we will choose only one

So, no. of possible outcomes from option C :

^5C_1\times ^6C_1

\frac{5!}{1!(5-1)!} \times \frac{6!}{1!(6-1)!}

\frac{5!}{1!(4)!} \times \frac{6!}{1!(5)!}

5 \times 6

30

Thus no. of possible outcomes from Option C is 30.

So, Option C is correct

Option D:  Select a number from {1, 2, 3, 4, 5} and a consonant

Since we are given that the code contains one number and one letter

No.of consonants  = 21

Out of these twenty one we will choose only one

We are supposed to choose a number from {1, 2, 3, 4,5}

Out of these five numbers we will choose only one

So, no. of possible outcomes from option C :

^5C_1\times ^21C_1

\frac{5!}{1!(5-1)!} \times \frac{21!}{1!(21-1)!}

\frac{5!}{1!(4)!} \times \frac{21!}{1!(20)!}

5 \times 21

105

Thus no. of possible outcomes from Option D is 105.

Hence Option C is correct because no. of possible outcomes in that case is 30 .

So, the code should have the structure C.

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This means that the data vary.  They are not the same data.  For example, "How old is Monday?" is not a statistical question because it has one answer.

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Answer:

The correct option is 1 - [(0.8)¹⁰+10*0.2*(0.8)⁹]= 0.6242

Step-by-step explanation:

Hello!

Given the distribution of probabilities for blood types for African-Americans:

O: 0.4

A: 0.2

B: 0.32

AB: 0.08

A random sample of 10 African-American is chosen, what is the probability that 2 or more of them have Type A blood?

Let X represent "Number of African-Americans with Type A blood in a sample of 10.

Then you have two possible outcomes,

"Success" the person selected has Type A blood, with an associated probability p= 0.2

"Failure" the selected person doesn't have Type A blood, with an associated probability q= 0.8

(You can calculate it as "1-p" or adding all associated probabilities of the remaining blood types: 0.4+0.32+0.08)

Considering, that there is a fixed number of trials n=10, with only two possible outcomes: success and failure. Each experimental unit is independent of the rest and the probability of success remains constant p=0.2, you can say that this variable has a Binomial distribution:

X~Bi(n;p)

You can symbolize the asked probability as:

P(X≥2)

This expression includes the probabilities: X=2, X=3, X=4, X=5, X=6, X=7, X=8, X=9, X=10

And it's equal to

1 - P(X<2)

Where only the probabilities of X=0 and X=1 are included.

There are two ways of calculating this probability:

1) Using the formula:

P(X)= \frac{n!}{(n-X)!X!} *p^{x} * q^{n-x}

With this formula, you can calculate the point probability for each value of X=x₀ ∀ x₀=1, 2, 3, 4, 5, 6, 7, 8, 9, 10

So to reach the asked probability you can:

a) Calculate all probabilities included in the expression and add them:

P(X≥2)= P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + X=10

b) Use the complement rule and calculate only two probabilities:

1 - P(X<2)= 1 - [P(X=0)+P(X=1)]

2) Using the tables of the binomial distribution.

These tables have the cumulative probabilities listed for n: P(X≤x₀)

Using the number of trials, the probability of success, and the expected value of X you can directly attain the corresponding cumulative probability without making any calculations.

>Since you are allowed to use the complement rule I'll show you how to calculate the probability using the formula:

P(X≥2) = 1 - P(X<2)= 1 - [P(X=0)+P(X=1)] ⇒

P(X=0)= \frac{10!}{(10-)0!0!} *0.2^{0} * 0.8^{10-0}= 0.1074

P(X=1)= \frac{10!}{(10-1)!1!} *0.2^{1} * 0.8^{10-1}= 0.2684

⇒ 1 - (0.1074+0.2684)= 0.6242

*-*

Using the table:

P(X≥2) = 1 - P(X<2)= 1 - P(X≤1)

You look in the corresponding table of n=10 p=0.2 for P(X≤1)= 0.3758

1 - P(X≤1)= 1 - 0.3758= 0.6242

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Full text in attachment.

I hope it helps!

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