For each device, there are only two possible outcomes. Either it fails, or it does not fail. The probability of a device failling is independent of other devices. 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.

A total of 15 devices will be used.

This means that $n = 15$

Assume that each device has a probability of 0.05 of failure during the course of the monitoring period.

This means that $p = 0.05$

What is the probability that one of the devices fail?

This is $P(X = 1)$

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

$P(X = 1) = C_{15,1}.(0.05)^{1}.(0.95)^{14} = 0.3658$

36.58% probability that one of the devices fail

7 0

2 years ago

HURRY I'LL PICK A BRAINLIEST PERSON!!!

mafiozo [28]

Hi!

I believe the answer is D , because the common denominator is 12 .

3 times 4=12
4 times 3=12
2 times 6= 12

You multiply each fraction by 12/1 .

Hope it helps and have a wonderful day !

5 0

2 years ago

Read 2 more answers

Value of m= 2 3m square - 2m - 7

Alex Ar [27]

Answer:

11m

Step-by-step explanation:

6 0

3 years ago