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.

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 bywhere
Here we're given
p=0.10  [ success = defective ]
n=3

(a) x=0





(b) x=2





(c) x ≥ 2

She should use 12/3 to measure

Igot the answer of 15.8 years.