Question

The firearms manufacturer that sells its firearms to the local sheriff’s department has 2000 used firearms to sell. Each used firearm has a 13% probability of needing major mechanical repairs. You randomly select 600 firearms for potential purchase.

What is the likelihood that 39 firearms will need major mechanical repairs? [Show your work]

What is the likelihood that all the firearms will work? [Show your work]

What is the likelihood that all the firearms will need major mechanical repairs? [Show your work]

Answer 1

Please find the answer as given below:

Given:

Probability that firearm need major mechanical repair = 0.13

Probability that firearm not need major mechanical repair =1 - 0.13 = 0.87

In given problem we have only two possible outcomes:

firearm need major mechanical repair and firearm not need major mechanical repair .

If we have only two possible outcomes then we use Binomial probability distribution to find out probability.

Formula for Binomial probability distribution:

P(X) =     

Where,

n = fixed number of trials,

x= total number of success out of n trials.

p = Probability of success

q = probability of failure.

p + q = 1, p and q are always between 0 to 1.

In binomial trial only two outcomes are possible. Success and failure.

Probability of success + Probability of failure = 1.

Solution 1)

Given: n = 600 , x = 39 , p = 0.13 and q = 1- p   = 1-0.13 = 0.87 ; q= 0.87

Put these values in binomial distribution formula:

P(X = 39 )

                                                                              Formula for combination nx =

=! *

= 0.000001 that is approximately zero.

Conclusion: probability of 39 firearm need to repair is zero.

Answer 2)

Given: n = 600 , x = 600 , p = 0.87 and q = 1- p   = 1-0.87 = 0.13 ; q= 0.13

Put these values in binomial distribution formula:

P(X = 600 )

                                                                              Formula for combination nx = n!x!n-x!

= *(

= 0.000 that is zero.

Conclusion: probability of 600 firearm need not to repair is zero.

Answer 3)

Given: n = 600 , x = 600 , p = 0.13 and q = 1- p   = 1-0.13 = 0.87 ; q= 0.87

Put these values in binomial distribution formula:

P(X = 600 )

                                                                              Formula for combination nx = n!x!n-x!

= *

= 0.000 that is zero.

Conclusion: probability of 600 firearm need to repair is zero.