Question

A manufacturer of a consumer electronics products expects 0.03 of units to fail during the warranty period. A sample of 500 independent units is tracked for warranty performance.(a) What is the probability that none fail during the warranty period?(b) What is the expected number of failures during the warranty period?(c) What is the probability that more than 2 units fail during the warranty period?Round your answer to 4 decimal places

Answer 1

a)
`P(X=0) = 500C0 (0.03)^0 (1-0.03)^(500-0) = 2.43x10^(-7)`

b)
`E(X)= n*p = 500*0.03=15`

c)
`P(X>2) = 1-P(X\leq 2)=0.99996`

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=500, p=0.03)`

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=(n!)/((n-x)! x!)`

Part a

We want this probability:

`P(X=0)`

Using the probability mass function we got:

`P(X=0) = 500C0 (0.03)^0 (1-0.03)^(500-0) = 2.43x10^(-7)`

Part b

The expected number are:

`E(X)= n*p = 500*0.03=15`

Part c

We want this probability:

`P(X>2) = 1-P(X\leq 2)`

And we can find the individual probabilities and we got:

`P(X=0) = 500C0 (0.03)^0 (1-0.03)^(500-0) = 2.43x10^(-7)`

`P(X=1) = 500C1 (0.03)^1 (1-0.03)^(500-1) = 3.76x10^(-6)`

`P(X=2) = 500C2 (0.03)^2 (1-0.03)^(500-2) = 2.90x10^(-5)`

And after replace we got:

`P(X>2) = 1-P(X\leq 2)=0.99996`