For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

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

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

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

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

And p is the probability of X happening.

The probability of a defective calculator is 10 percent.

This means that $p = 0.1$

If 3 calculators are selected at random, what is the probability that one of the calculators will be defective

This is P(X = 1) when n = 3. So

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

$P(X = 1) = C_{3,1}.(0.1)^{3}.(0.9)^{2} = 0.243$

There is a 24.3% probability that one of the calculators will be defective.

3 0

3 years ago

Wat is the domain of g(x)

scoundrel [369]

Answer:

The first choice is the answer of course

3 0

2 years ago

A molasses can is made with a cylindrical base having a radius of 6 cm, that is topped by a smaller cylindrical pouring spout wi

defon

Answer:

Step-by-step explanation:

The total height of the can is 20cm

3 0

2 years ago

A randomly generated password contains four characters. Each of the four characters is either a lowercase letter or a digit from

juin [17]

Sample space is 36C4
Now, we want to know all of the combinations that have 1 digit in it.
So, we can have one here:

1XXX
X1XX
XX1X
XXX1

But we have 10 different digits to choose from. So, we need to introduce the combination term, nCr, where n is a list of all digits and r is how many we want.

Since we only want one, we will need 10C1 for the number of digits. But we need to choose three lowercases, so it becomes 10C1 × 26C3

Since it's a probability question, we need to divide that by our sample space, 36C4, and our percentage becomes 44%

6 0

3 years ago

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