Question

A sequence of Bernoulli trials consists of choosing components at random from a batch of components. A selected component is either classified as defective or nondefective. If the probability that a selected component is non-defective is 0.8, find the following probabilities: a) Three non-defective components in a batch of seven components. b) 8 non-defective components are drawn before the first defective component is chosen.

Answer 1

Answer:

a) 0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.

Step-by-step explanation:

A sequence of Bernoulli trials composes the binomial distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that a selected component is non-defective is 0.8

This means that $p = 0.8$

a) Three non-defective components in a batch of seven components.

This is P(X = 3) when n = 7. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{7,3}.(0.8)^{3}.(0.2)^{4} = 0.0287$

0.0287 = 2.87% probability that three non-defective components in a batch of seven components.

b) 8 non-defective components are drawn before the first defective component is chosen.

Now the order is important, so the we just multiply the probabilities.

8 non-defective, each with probability 0.8, and then a defective, with probability 0.2. So

$p = (0.8)^8*0.2 = 0.0336$

0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.