Question

A machine has 12 identical components which function independently. The probability that a component will fail is 0.2. The machine will stop working if more than three components fail. Find the probability that the machine will stop.

Answer 1

Answer:

$P(X>3) =1- P(X\leq 3) = 1-[P(X=0)+P(X=1) +P(X=2) +P(X=3)]$

And if we find the individual probabilities we got:

$P(X=0)=(12C0)(0.2)^0 (1-0.2)^{12-0}=0.0687$

$P(X=1)=(12C1)(0.2)^1 (1-0.2)^{12-1}=0.2062$

$P(X=2)=(12C2)(0.2)^2 (1-0.2)^{12-2}=0.2835$

$P(X=3)=(12C3)(0.2)^3 (1-0.2)^{12-3}=0.2362$

And if we replace we got:

$P(X>3) =1- P(X\leq 3) = 1-[0.0687+0.2062 +0.2835 +0.2362]=0.20543$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=12, p=0.2)$

And we want to find this probability:

$P(X > 3)$

And we can use the complement rule and we got:

$P(X>3) =1- P(X\leq 3) = 1-[P(X=0)+P(X=1) +P(X=2) +P(X=3)]$

And if we find the individual probabilities we got:

$P(X=0)=(12C0)(0.2)^0 (1-0.2)^{12-0}=0.0687$

$P(X=1)=(12C1)(0.2)^1 (1-0.2)^{12-1}=0.2062$

$P(X=2)=(12C2)(0.2)^2 (1-0.2)^{12-2}=0.2835$

$P(X=3)=(12C3)(0.2)^3 (1-0.2)^{12-3}=0.2362$

And if we replace we got:

$P(X>3) =1- P(X\leq 3) = 1-[0.0687+0.2062 +0.2835 +0.2362]=0.20543$

Answer 2

Answer: Answer = 0.275

Step-by-step explanation:

Given p= 0.2, q = 1 - 0.2 = 0.8

We apply the Bernoulli probability distribution.

P(X=r) = nCr (p)^r (q)^n-r

n= 12(since components are independent)

Let probability that machine stops working be P(X), where is the component. Then

P(X)= (X>3) = 1 - P(X<or=3)

Since machine stop working at X>3

So for P(X< or =3) we consider r= 1, 2, 3.

Hence P(X>3) = 1 - [P(X=r=1) + P(X=r=2) + P(X=r=3)]

P(X>3) = 12C1 (0.2)¹(0.8)¹¹ + 12C2(0.2)²(0.8)^10 +12C3(0.2)³(0.8)^9

P(X>3) = 1 - [0.206 + 0.283 + 0.236 ]

P(X>3) = 1 - 0.725

P(X>3) = 0.275