Question

Your company inspects car tires before final shipment. tires are judged to be either defective, or non-defective. The probability of a tire being found defective is 6%

1. What is the probability it will take 3 tests or fewer to find 2 defective tires?
2. What is the probability it takes more than 2 tests to find the first defective tire?
3. If the inspector randomly selects 30 tires from the holding area, what is the probability 4 tires will be found defective?

Answer 1

Probability of a tire being found defective = 6 % = 0.06

We define a discrete random variable X = no. of tires found defective .

We can create a Binomial distribution for X considering each test as a trial Because:

(i) no of trials would be some fixed value n

(ii) there are only two possible outcomes for each trial

(iii) probability of success is independent for each trial

no. of trial = n

Probability of Success ( we consider tire being defective as success) = p = 0.06

So X ~ B(n,p)

and Probability distribution is given as :

  

So B(n,p,x) = Probability of getting 'x' defective tires in 'n' no. of tests where probability of success = p

.

a) We need probability that it will take 3 or fewer tests to find 2 defective tires

We know that it will take atleast 2 tests to find 2 defective tires. (we can't find two defective if we have only tested one )

So required probability = Probability that it will take 2 or 3 fewer test to find 2 defective .

= Probability that it will take 2 test + Probability that it will take 3 tests to find 2 defective tires

Probability that we find 2 defective in 2 test = B(n,p,x)

  

Probability that it takes 3 tests to find 2 defective = for this we need to make sure that the 2nd defective tire is found on the 3rd test only

=Probability that we find 1 defective in first 2 tests and 1 defective tire in 3rd test

= B(2, 0.06 , 1) * B(1 , 0.06 , 1)

= 0.1128 * 0.06 = 0.0068

Required probability = 0.0036 + 0.0068 = 0.0104

b.  Probability that it takes more than 2 tests to find first defective tire =

= Probability that we find no defective tires in first 2 tests

Required probability   

c. If total 30 tires are tested , meaning n = 30

We need probability that 4 tires will be found defective = P(X=4 ) = B(30 , 0.06 , 4)