Question

2. A company manufactures fuses. The percentage of non-defective fuses is 95.4%. A sample of 9 fuse was selected. Calculate the probability of selecting at least 3 defective fuses.

Answer 1

Answer:

0.0067 = 0.67% probability of selecting at least 3 defective fuses.

Step-by-step explanation:

For each fuse, there are only two possible outcomes. Either it is defective, or it is not. The probability of a fuse being defective is independent of any other fuse, which means that the binomial probability distribution is used to solve this question.

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.

A company manufactures fuses. The percentage of non-defective fuses is 95.4%.

This means that 100 - 95.4 = 4.6% = 0.046 are defective, which means that $p = 0.046$

A sample of 9 fuse was selected.

This means that $n = 9$

Calculate the probability of selecting at least 3 defective fuses.

This is:

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{9,0}.(0.046)^{0}.(0.954)^{9} = 0.6545$

$P(X = 1) = C_{9,1}.(0.046)^{1}.(0.954)^{8} = 0.2840$

$P(X = 2) = C_{9,2}.(0.046)^{2}.(0.954)^{7} = 0.0548$

Then

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.6545 + 0.2840 + 0.0548 = 0.9933$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.9933 = 0.0067$

0.0067 = 0.67% probability of selecting at least 3 defective fuses.