Question

A factory makes 10% defective items and items are independently defective. If a sample of 10 items is to be selected, find the probability that 9 or more are NOT defective in two ways. (Round to 3 decimal places) g

Answer 1

Answer:

$P(x \geq 9)=P(X=9)+P(X=10)$

$P(X=9)=(10C9)(0.9)^9 (1-0.9)^{10-9}=0.387$

$P(X=10)=(10C10)(0.9)^{10} (1-0.9)^{10-10}=0.349$

And adding we got:

$P(x \geq 9)=P(X=9)+P(X=10)=0.387+0.349=0.736$

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

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

$X \sim Binom(n=10, p=1-0.1=0.9)$

Where 1-p = 1-0.1=0.9 represent the probability of being NOT defective

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:

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

And we want to find this probability:

$P(x \geq 9)=P(X=9)+P(X=10)$

$P(X=9)=(10C9)(0.9)^9 (1-0.9)^{10-9}=0.387$

$P(X=10)=(10C10)(0.9)^{10} (1-0.9)^{10-10}=0.349$

And adding we got:

$P(x \geq 9)=P(X=9)+P(X=10)=0.387+0.349=0.736$