The quotient of 6 divided by 12

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

Elis [28]

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]

7 0

4 years ago

Help asap will mark brainliest

Roman55 [17]

Answer:

74

Step-by-step explanation:

Both angles x and 106 are supplementary which means

x+106=180

x=180-106

x=74

8 0

3 years ago

Read 2 more answers

What is this equations name: ax+by=c

Tju [1.3M]

This is standard form for a linear equation

4 0

3 years ago

An electronic product contains 40 integrated circuits. The probability that any integrated circuit is defective is 0.01, and the

neonofarm [45]

Answer:

There is a 33.10% probability that there is at least one defective integrated circuit.

Step-by-step explanation:

For either integrated circuit, there are only two possible outcomes. Either they are defective, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

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

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem

The electonic product contains 40 integrated circuits, so $n = 40$ .

The probability that any integrated circuit is defective is 0.01, so $\pi = 0.01$

What is the probability that there is at least one defective integrated circuit?

Either there is at least one defective integrated circuit, that is probability $P(X > 0)$ , or there are no defective integrated circuits, that is probability $P(X = 0)$ . The sum of these probabilities is decimal 1. We want to find $P(X>0)$ .