Ten percent of the engines manufactured on an assembly line are defective. a. If engines are randomly selected one at a time and tested what is the probability that the first non defective engine will be found on the second trial?
b. What is the probability that the third non defective engine will be found on the fifth trial?
c. Find the mean and variance of the number of trials on which the first non defective engine is found.
d. Find the mean and variance of the number of trials on which the third non defective engine is found.

a) the probability that the first non-defective engine will be found on the second trial is 0.09

b) probability that the third non-defective engine will be found on the fifth trial is 0.00486

c) the Mean is 1.1111  and Variance is 0.1235  

d) the Mean is 3.3333and Variance is 0.3704  

Step-by-step explanation:

Firstly;

Let x be the number of trail on which the rth defective occurs

Also let the probability that an occurrence of a defective be p = 0.10

Here, x follows negative binomial distribution with parameters r and p

The probability mass function of X is as follows:

P(X = x) = [ x -1  p^r ( 1 - p)^(x-r)       ; x = r, r + 1, r + 2, ...

                 r - 1 ]

=   [ x -1  (0.10)^r ( 1 - 0.10)^(x-r)

     r - 1 ]

= [ x -1  (0.10)^r ( 0.9)^(x-r)  n 1 let this be equatio

    r - 1 ]

This represents the probability that the rth success occurs on the xth trail.  

a)

probability that the first non-defective engine will be found on the second trial?

Substitute r = 1 and x = 2 in equation 1  

P(X = 2)  = [ 2 - 1   (0.10)¹ ( 0.90 )²⁻¹

                   1 - 1 ]

= (0.10) × (0.90)

= 0.09

Therefore, the probability that the first non-defective engine will be found on the second trial is 0.09

b)

probability that the third non-defective engine will be found on the fifth trial?

So we substitute r = 3 and x = 5 in equation 1  

P(X = 5) = [ 5 - 1  (0.10)³ ( 0.90)⁵⁻³

                  3 - 1 ]

=    [ 4    (0.10)³ ( 0.9)²

      2 ]

= 0.00486

Therefore, probability that the third non-defective engine will be found on the fifth trial is 0.00486

Now the formula for the mean of the negative binomial distribution is as follows:  

Mean u = r / p     let this be equation 2

The formula for the variance of the negative binomial distribution also is as follows:  

Variance α² = rq / p²   let this be equation 3

so

c)

the mean and variance of the number of trials on which the first non-defective engine is found.  

First, let the probability that non-defective engine found be p = 0.90

And q = (1 - p) = 1 - 0.90 = 0.10  

Now we substitute r = 1, p = 0.90 and q = 0.10 in equation 2 & 3simultaneously,

the mean and variances are as follows;

Mean = r/p = 1/0.90 = 1.1111

Variance = rq/p² = (1)(0.10) / (0.90)² = 0.1235  

Therefore the Mean is 1.1111  and Variance is 0.1235  

d)

the mean and variance of the number of trials on which the third non-defective engine is found  

Let the probability that non-defective engine found be p = 0.90

And q = (1 - p) = 1 - 0.90 = 0.10

Now we substitute r = 3, p = 0.90 and q = 0.10 in equation (2) and (3) simultaneously,

the mean and variances are;

Mean = r/p = 3/0.90 = 3.3333

Variance = rq/p² = (3)(0.10) / (0.90)² = 0.3704

Therefore the Mean is 3.3333 and Variance is 0.3704