Please help due in 10 mins
1 answer:
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Answer:
a) 96.04% probability of getting no defectives.
b) 3.92% probability of getting 1 defective.
c) 0.04% probability of getting 2 defectives.
You pick 2 plugs, so either you get no defective, 1 defective, or both defective. The sum of these probabilities must be 100%, which is what we get.
Step-by-step explanation:
For each plug, there are only two possible outcomes. Either it is defective, or it is not. The probability of a plug being defective is independent of other plugs. So we use the binomial probability distribution 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.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
2 % over a long time and that this process is controlled every half hour by drawing and inspecting two just produced.
This means that
(a) no defectives,
P(X = 0)
96.04% probability of getting no defectives.
(b)1 defective,
P(X = 1)
3.92% probability of getting 1 defective.
(c) 2 defectives.
P(X = 2)
0.04% probability of getting 2 defectives.
What is the sum of these probabilities?
You pick 2 plugs, so either you get no defective, 1 defective, or both defective. The sum of these probabilities must be 100%, which is what we get.