For each item, there are only two possible outcomes. Either they are defective, or they are not. Items are independent of each other. So 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.

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

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.

5 percent of these are defective.

This means that $p = 0.05$

If two items are randomly selected as they come off the production line, what is the probability that both are defective

This is P(X = 2) when n = 2. So

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

$P(X = 2) = C_{2,2}.(0.05)^{2}.(0.95)^{0} = 0.0025$

0.0025 = 0.25% probability that both are defective

4 0

2 years ago

Six hours more per week then Theodore studies

Ierofanga [76]

Answer:

umm you have to give us more info

Step-by-step explanation:

without info we can not properly answer the question

5 0

3 years ago

Three fourths of a number p less than two is five sixths of another number r plus five solve for r

maw [93]

r = -3/4p - 18/5
(or you could put 3.6 or 3 3/5 instead of 18/5)

2 - 3/4p = 5/6r + 5
-5 -5
-3 - 3/4p = 5/6r
/(5/6) /(5/6)
-18/5 - 3/4p = r
-3/4p - 18/5 = r

5 0

3 years ago