Question

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is required to do three welds in a day:

a. what is the probability that none of welds will be defective?
b. what is the probability that exactly two of welds will be defective? 2
c. what is the probability that at least two welds are defective? assume that the condition of each weld is independent of the condition of the other welds.

Answer 1

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by
`P(x)=C(n,x)p^x(1-p)^(n-x)`where
`C(n,x)=(n!)/(x!(n-x)!)`
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0

`P(x)=C(n,x)p^x(1-p)^(n-x)`

`=C(3,0)0.1^0(1-0.1)^(3-0)`

`=1(1)(0.729)`

`=0.729`

(b) x=2

`P(x)=C(n,x)p^x(1-p)^(n-x)`

`=C(3,2)0.1^2(1-0.1)^(3-2)`

`=3(0.01)(0.9)`

`=0.027`

(c) x ≥ 2

`P(x)=\sum_(x=2)^3C(n,x)p^x(1-p)^(n-x)`

`=P(2)+P(3)`

`=C(n,2)p^2(1-p)^(n-2)+C(n,3)p^3(1-p)^(n-3)`

`=C(3,2)0.1^2(1-0.1)^(3-2)+C(3,3)0.1^3(1-0.1)^(3-3)`

`=3(0.01)(0.9)+1(0.001)1`

`=0.027+0.001`

`=0.028`