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.

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

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