chanavictor2688
02.12.2020 •
Mathematics
A certain manufacturing plant produces electric fuses of which 20% are defective. Find the probability that in a sample of 8 fuses selected at random there will be atleast one defective and not more than one defective.
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Ответ:
0.419
Step-by-step explanation:
To calculate this, we shall make use of Bernoulli approximation of Binomial distribution
if 20% are defective, then 80% are not defective
Probability of selecting a defective fuse is 20/100 = 0.2
Probability of selecting a non defective one is 0.8
Probability of at least 1 being defective is = 1 - Probability of none being defective
Mathematically that will be;
0.8^8 = 0.168
The above is probability of none defective
So the probability of at least one will be
1 - 0.168 = 0.832
Probability of not more than 1 means;
Probability of none + probability of 1 being defective
We already have probability of none above
Probability of 1 being defective means 8 will be non defective
The probability in this case is;
8 C 1 0.2^1 0.8^7
= 8 * 0.2 * 0.8^7 = 0.336
Add this to the probability of none = 0.336 + 0.168 = 0.504
So the probability that we want to calculate from the question will be;
Probability of at least one defective * probability of not more than one defective
= 0.832 * 0.504 = 0.419
Ответ:
hope this helps you