Question

Ten percent of the engines manufactured on an assembly line are defective. If engines are randomly selected one at a time and tested, what is the probability the fourth defective engine will be found on the second trial?

Answer 1

The probability the fourth defective engine will be found on the second trial is 0.0001 or 0.01%.

Step-by-step explanation:

To find the probability that the fourth defective engine will be found on the second trial, we need to consider the probability of finding three defective engines in the first trial and then finding the fourth defective engine in the second trial. Since 10% of the engines are defective, the probability of finding a defective engine in one trial is 0.10.

So, the probability of finding three defective engines in the first trial is (0.10)^3 = 0.001.

Then, the probability of finding the fourth defective engine in the second trial is 0.10.

The overall probability is the product of these two probabilities: 0.001 * 0.10 = 0.0001 or 0.01%.

Answer 2

0

Step-by-step explanation:

From the given information;

Let x be the number of trails on which the
`r^(th)` defective occurs.

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

Suppose x is a random variable that follows a negative binomial distribution with parameters r and p.

Then the probability mass function of X can be expressed as:

`P(X=x) = \bigg (^(x-1)_(r-1) \bigg)p^r (1 - p) ^(x-r) \ \ \ \ ; x=r, \ r+1 , \ r+2 , ...`

`P(X=x) = \bigg (^(x-1)_(r-1) \bigg)(0.10)^r (1 - 0.10) ^(x-r)`

`P(X=x) = \bigg (^(x-1)_(r-1) \bigg)(0.10)^r ( 0.90) ^(x-r) \ \ ... \ \ (1)`

We are to find the probability of the fourth defective engine will be found on the second trial.

i.e. r = 4 and x = 2

`P(X=2) = \bigg (^(2-1)_(4-1) \bigg)(0.10)^4 ( 0.90) ^(2-4)`

`P(X=2) = \bigg (^1}_(3) \bigg)(0.10)^4 ( 0.90) ^(-2)`

`P(X=2) = \bigg ((1!)/(3!(1-3)!) \bigg)(0.10)^4 ( 0.90) ^(-2)`

`\mathbf{P(X=2) = 0}`