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muminat
4 years ago
5

A class with n kids lines up for recess. The order in which the kids line up is random with eachordering being equally likely. T

here are three kids in the class named John, Betty, and Mary. The useof the word "or" in the description of the events, should be interpreted as the inclusive or. That is"A or B" means thatAis true,Bis true, or bothAandBare true.Give an expression for each of the probabilities below as a function ofn. Simplify your final expressionas much as possible so that your answer does not include any expressions of the form (a/b).Requried:a. What is the probability that Betty is first in line or Mary is last in line? b. Explain the method used to calculate probability.
Mathematics
1 answer:
Elis [28]4 years ago
7 0

Answer:

A) P(Betty is first in line and mary is last) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))

B) The method used is Relative frequency approach.

Step-by-step explanation:

From the question, we are told a sample of n kids line up for recess.

Now, the order in which they line up is random with each ordering being equally likely. Thus, this means that the probability of each kid to take a position is n(total of kids/positions).

Since we are being asked about 3 kids from the class, let's assign a letter to each kid:

J: John

B: Betty

M: Mary

A) Now, we want to find the probability that Betty is first in line or Mary is last in line.

In this case, the events are not mutually exclusive, since it's possible that "Betty is first but Mary is not last" or "Mary is last but Betty is not first" or "Betty is the first in line and Mary is last". Thus, there is an intersection between them and the probability is symbolized as;

P(B₁ ∪ Mₙ) = P(B₁) + P (Mₙ) - P(B₁ ∩ Mₙ) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))

Where;

The suffix 1 refers to the first position while the suffix n refers to the last position.

Also, P(B₁ ∩ Mₙ) = P(B₁) × P(Mₙ/B₁)

This is because the events "Betty" and "Mary" are not independent since every time a kid takes his place the probability of the next one is affected.

B) The method used is Relative frequency approach.

In this method, the probabilities are usually assigned on the basis of experimentation or historical data.

For example, If A is an event we are considering, and we assume that we have performed the same experiment n times so that n is the number of times A could have occurred.

Also, let n_A be the number of times that A did occur.

Now, the relative frequency would be written as (n_A)/n.

Thus, in this method, we will define P(A) as:

P(A) = lim:n→∞[(n_A)/n]

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<u>ANSWER:</u><u> </u>
The original number is 64

<u>SOLUTION:</u>
Let the two-digit number be ab, where 'a' is the tens digit and 'b' is the units digit.
<span>"The sum of the digits of a two-digit number is 10" implies
a + b = 10 (Equation 1)

</span><span>"When the digits are reversed, the new number is 18 less than the original number" is algebraically written as
ba = ab - 18
>> bx10 + ax1 = (ax10 + bx1) - 18
</span>>> 10b + a = 10a + b - 18
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Now substitute equation 2 in equation 1 to get
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>> 2b = 8
>> b = 4

Plug this value back in equation 1 to get
a = b + 2 = 4 + 2 = 6

So a = 6 and b = 4. Hence, the original number is 64.

<u>CHECK</u><u>:</u>
<span><u>Given</u>: The sum of the digits of a two-digit number is 10
</span><u>Check</u>: 6 + 4 = 10 (so it is true)

<span><u>Given</u>: When the digits are reversed, the new number is 18 less than the original number
</span><u>Check</u>:
Digits reversed: 64 reversed becomes new number 46, and 46 is 18 less than 64, that is, 46 = 64 - 18. 



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